Answer:
c. (CH3CH2CH2)2CuLi
Explanation:
(CH3CH2CH2)2CuLi
This reagent is called Lithium di(n propyl)cuprate used to covert benzoyl chloride to phenyl propyl ketone. this reagent is called Gilman Reagent and The reaction occur with nucleophilic substitution of an alkyl group for the leaving group (chloride), forming one new carbon-carbon bond.
Hence the correct answer is C that is c. (CH3CH2CH2)2CuLi
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X , the freezing point of the solution is 2.7°C lower than the freezing point of pure X. On the other hand, when 70.4 g of ammonium chloride (NH4CI) are dissolved in the same mass of X, the freezing point of the solution is 9.9 °C lower than the freezing point of pure X.
A) Calculate the van't Hoff factor for ammonium chloride in X. Be sure your answer has a unit symbol, if necessary, and round your answer to 2 significant digits.
Answer:
i=1.62 .
Explanation:
Let, i be the Van't Hoff Factor.
Moles of benzamide,=[tex]\dfrac{Mass}{molar\ mass}=\dfrac{70.4}{121.14}=0.58\ mol.[/tex]
Molality of solution, m=[tex]\dfrac{moles }{mass\ of\ solvent}=\dfrac{0.58}{0.85}=0.68\ molal.[/tex]
Now, we know
Depression in freezing point, [tex]\Delta T=i\times K_f\times m[/tex] .....1
It is given that,
[tex]\Delta T=2.7^o C\\i=1 ( since\ it\ is\ non\ dissociable\ solutes)\\K_f ( freezing\ constant)\\[/tex]
Putting all these values we get,
[tex]K_f=3.949\ C/m.[/tex]
Now, moles of ammonium chloride=[tex]\dfrac{70.4}{53.49}=1.316\ mol.[/tex]
molality =[tex]\dfrac{1.316}{0.85}=1.54 molal.\\\Delta T=9.9 .[/tex]
Putting all these values in eqn 1.
We get,
i=1.62 .
Hence, this is the required solution.
1. Explain the difference between a chemical change and a physical change.
2. List three examples of physical changes and three examples of chemical changes. Explain why these are chemical or physical changes.
3.If you were boil water in a pot.Chemical changes:
1.Your food digesting.
2.Cooking an egg for breakfast.
3.The milk souring in your fridge.
Answer:
1. The difference between a chemical change and a physical change is as follows;
In a physical change no new substance is formed while in a chemical change new substance are formed
Chemical change involves great amount of heat changes while physical change does not involves great amount of heat change except the latent heat changes which occur during changes of state
Physical change is easily reversed while a chemical change is not easily reversed
There is no change in the mass of a substance involve in physical change while in chemical change there is a change in the mass of a substance that undergoes such a change
2 . examples of physical and chemical changes
Physical change.
1. The magnetization and demagnetization of iron rods( the iron can be removed and no new substance will be formed)
11. The separation of mixtures by evaporation, distillation, fractional distillation, sublimation.( the separation techniques mentioned are physical change because they can be easily revised to get the original quantity back, like when water evaporates to vapour it can be retrieved back by cooling)
111. Changes in the states of matter such as melting of solids to liquids the freezing of liquids to solids, the vaporization of liquids to gases.( also changes of states are physical change because no new substance will be formed and it is easily reversible)
Chemical change
1. Fermentation and decay of substances .( new substance is formed which is the decayed substance)
11. The rusting of iron( rusting of iron is a chemical change because a new substance is formed. Which is the rusted iron)
111. The addition of water to quick lime I.e the slacking of lime. ( is a chemical change because a new substance is formed)
3. Boiling of water in a pot is a physical change because the water will evaporate into vapour at 100° Celsius and can be regained by cooling
Food digesting is a chemical change because a new substance will be formed , which is faeces.
Cooking an egg for breakfast is a chemical change because a new substance will be formed which is boiled egg and it can not be reversed.
Souring of milk is a chemical change because a new substance will be formed which is the soured taste
1. Chemical changes alter a substance's composition (e.g., burning wood), while physical changes affect its physical properties (e.g., melting ice).
2. Examples: Physical (melting ice), Chemical (burning wood).
3. Boiling water is a physical change; food digestion, cooking an egg, and milk souring are chemical changes due to altered composition.
1. **Difference between Chemical and Physical Changes**:
- **Chemical Change**: A chemical change involves the alteration of a substance's chemical composition. New substances with different properties are formed. This change is typically irreversible and accompanied by a release or absorption of energy.
- **Physical Change**: A physical change, on the other hand, does not alter the substance's chemical identity. It involves changes in physical properties, such as size, shape, phase (solid, liquid, gas), or state (e.g., melting, freezing), but the chemical composition remains the same. Physical changes are generally reversible, and no new substances are created.
2. **Examples of Physical and Chemical Changes**:
**Physical Changes**:
- **Ice Melting**: When ice (solid water) melts to form liquid water, it undergoes a physical change. The chemical composition of H₂O remains the same, but the phase changes from solid to liquid.
- **Cutting Paper**: When you cut a piece of paper into smaller pieces, it's a physical change. The chemical composition of cellulose in the paper remains unchanged.
- **Boiling Water**: Boiling water represents a physical change. Water changes from a liquid to a gas (steam) without any alteration in its chemical composition.
**Chemical Changes**:
- **Burning Wood**: When wood is burned, it undergoes a chemical change. The cellulose and other compounds in wood react with oxygen to produce carbon dioxide and water vapor. New substances are formed, and energy is released.
- **Digesting Food**: The process of digesting food in your body involves chemical changes. Enzymes break down complex molecules into simpler ones, allowing your body to absorb nutrients.
- **Milk Souring**: When milk sours, it undergoes a chemical change due to the action of bacteria. Lactose in milk is converted into lactic acid, changing the taste and composition of the milk.
3. **Boiling Water - Physical Change**:
Boiling water is a physical change because it involves a change in the physical state of water from liquid to gas (steam), but the chemical composition remains H₂O.
**Chemical Changes**:
1. **Your Food Digesting**: Digestion involves chemical changes as enzymes break down complex food molecules into simpler ones, which can then be absorbed by your body.
2. **Cooking an Egg for Breakfast**: Cooking an egg is a chemical change because heat causes proteins in the egg white to denature and coagulate, forming a solid structure with different properties than the raw egg white.
3. **The Milk Souring in Your Fridge**: Milk souring is a chemical change. Bacteria in the milk metabolize lactose into lactic acid, changing the taste and composition of the milk.
In summary, chemical changes involve a change in chemical composition, while physical changes involve changes in physical properties without altering the chemical identity of a substance. Boiling water is a physical change, while food digestion, cooking an egg, and milk souring are chemical changes due to alterations in chemical composition and properties.
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The heat combustion of gaseous butane is -2658 kl/mol and the heat of combustion of liquid butane is:
Answer:
The heat of combustion of liquid butane is: -2636 kJ/mol
Explanation:
This problem is solved by using Hess law of heats summation:
We have the heat of combustion for gaseous butane:
C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l) ΔHºcomb = -2658 kJ/mol
and we want the heat of combustion for liquid butane.
What we need to do is add the heat of vaporization for liquid butane to this equation:
C₄H₁₀ (l) ⇒ C₄H₁₀ (g) ΔHºvap = 22.44 kJ/mol
C₄H₁₀ (g) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l) ΔHºcomb = -2658 kJ/mol
______________________________________________
C₄H₁₀ (l) + 13/2 O₂ (g) ⇒ 4 CO₂ (g) + H₂O (l)
So,
ΔH comb liq butane = -2658 kJ/mol + 22.44 kJ/mol = -2636 kJ/mol
Effective nuclear charge, Zeff, is defined as:
Zeff=Z−S
where Z is true nuclear charge and S is the amount of shielding.
In 1930, John C. Slater devised the following set of empirical rules to estimate S for a designated ns or np electron:
1. Write the electron configuration of the element, and group the subshells as follows: (1s), (2s, 2p), (3s, 3p), (3d), (4s, 4p), (4d), (4f ), (5s, 5p), and so on.
2. Electrons in groups to the right of the (ns, np) group contribute nothing to the shielding constant for the designated electron.
3. All the other electrons in the (ns, np) group shield the designated electron to the extent of 0.35 each.
4. All electrons in the n−1 shell shield to the extent of 0.85 each.
5. All electrons in the n−2 shell, or lower, shield completely—their contributions to the shielding constant are 1.00 each.
When the designated electron is in an nd or nf group, rules (i), (ii), and (iii) remain the same but rules (iv) and (v) are replaced by the following: Each electron in a group lying to the left of the nd or nf group contributes 1.00 to the shielding constant. These rules are a simplified generalization based on the average behavior of different types of electrons.
Part A) Calculate Zeff for a valence electron in an oxygen atom.
Answer:
The effective nuclear charge for a valence electron in oxygen atom: [tex]Z_{eff} = 4.55[/tex]
Explanation:
Effective nuclear charge [tex][Z_{eff}][/tex] is the net nuclear charge experienced by the electron in a given atom. It is always less than the actual charge of the nucleus [Z], due to shielding by electrons in the inner shells.
It is equal to the difference between the actual nuclear charge or the atomic number (Z) and the shielding constant (s).
[tex]\Rightarrow Z_{eff} = Z - s[/tex]
For an oxygen atom-
Electron configuration: (1s²) (2s² 2p⁴)
The atomic number (actual nuclear charge): Z = 8
The shielding constant (s) for a valence electron can be calculated by using the Slater's rules:
⇒ s = 5 × 0.35 + 2 × 0.85 = 1.75 + 1.7 = 3.45
Therefore, the effective nuclear charge for a valence electron in oxygen atom is:
[tex]Z_{eff} = Z - s = 8 - 3.45 = 4.55[/tex]
Therefore, the effective nuclear charge for a valence electron in oxygen atom: [tex]Z_{eff} = 4.55[/tex]
To calculate the effective nuclear charge (Zeff) for a valence electron in an oxygen atom, which has an atomic number of 8, we use Slater's rules to determine the shielding constant (S) and subtract it from Z. The result is Zeff = 8 - (2 * 0.85 + 5 * 0.35) = 4.55.
The effective nuclear charge (Zeff) for an electron can be determined using Slater's rules which consider the actual nuclear charge (Z) and the shielding constant (S). To calculate Zeff for a valence electron in an oxygen atom with an atomic number of 8:
Firstly, identify the electron configuration of oxygen which is 1s2 2s2 2p4.Group the subshells as Slater's rules suggest.Since we are interested in a valence electron in the 2p orbital, the two 1s electrons shield with a factor of 0.85 each, and the six electrons in the 2s and 2p orbitals (excluding the electron in question) shield with a factor of 0.35 each.Therefore, the shielding constant S is calculated as 2(0.85) + 5(0.35).The effective nuclear charge Zeff is then Z - S, which equals 8 - (2 * 0.85 + 5 * 0.35) = 4.55.The calculated Zeff for a valence electron in an oxygen atom is 4.55.
The great French chemist Antoine Lavoisier discovered the Law of Conservation of Mass in part by doing a famous experiment in 1775. In this experiment Lavoisier found that mercury(II) oxide, when heated, decomposed into liquid mercury and an invisible and previously unknown substance: oxygen gas.
1. Write a balanced chemical equation, including physical state symbols, for the decomposition of solid mercury(II) oxide (HgO) into liquid mercury and gaseous dioxygen.
2. Suppose 50.0ml of dioxygen gas are produced by this reaction, at a temperature of 90°C and pressure of exactly 1atm. Calculate the mass of mercury(II) oxide that must have reacted. Be sure your answer has the correct number of significant digits.
Answer:
(a) [tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]
(b) 0.726 g
Explanation:
(a)
The balanced chemical equation is shown below as:-
[tex]2HgO_{(s)}\rightarrow 2Hg_{(l)}+O_2_{(g)}[/tex]
(b)
Given that:
Temperature = 90 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (90 + 273.15) K = 363.15 K
V = 50.0 mL = 0.05 L ( 1 mL = 0.001 L )
Pressure = 1 atm
Using ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L atm/ K mol
Applying the equation as:
1 atm × 0.05 L = n ×0.0821 L atm/ K mol × 363.15 K
⇒n = 0.001677 mol
Thus, moles of [tex]O_2[/tex] = 0.001677 mol
According to the reaction shown above,
1 mole of [tex]O_2[/tex] is produced when 2 moles of mercury(II) oxide are reacted
0.001677 mole of [tex]O_2[/tex] is produced when [tex]2\times 0.001677[/tex] moles of mercury(II) oxide are reacted
Moles of mercury(II) oxide = 0.003354 moles
Molar mass of mercury(II) oxide = 216.59 g/mol
Mass of mercury(II) oxide = [tex]Moles\times Molar\ mass[/tex] = [tex]0.003354\times 216.59\ g=0.726\ g[/tex]
0.726 g is the mass of mercury(II) oxide that must have reacted.
The balanced chemical equation for the decomposition of mercury(II) oxide into mercury and dioxygen is: 2HgO(s) → 2Hg(l) + O2(g). The mass of mercury(II) oxide that reacted to produce 50.0 ml of dioxygen gas at 90°C and 1 atm pressure is approximately 0.97 g.
Explanation:1. The balanced chemical equation for the reaction is: 2HgO(s) → 2Hg(l) + O2(g).
2. First, we need to use the ideal gas law to calculate the number of moles of dioxygen gas produced. At standard temperature and pressure (STP), one mole of any gas occupies a volume of 22.4 L. Therefore, the number of moles of dioxygen is 50.0 ml / 22.4L/mol = 0.00223 moles. Next, we use the balanced chemical equation, which tells us that one mole of HgO produces a half mole of dioxygen gas. So, the amount of HgO reacted is 2 x 0.00223 moles = 0.00446 moles. Finally, we use the molar mass of HgO (216.6 g/mol) to calculate the mass of HgO reacted: 0.00446 moles x 216.6 g/mol = 0.97 g (to three significant digits).
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The ΔH°soln of HNO3 is –33.3 kJ/mol. 13.0 mL of 14.0 M HNO3 is dissolved in 100.0 mL of distilled water initially at 25°C. How much ice at 0°C [cP = 37.1 J/(mol ·°C), ΔH°fus = 6.01 kJ/mol] must be added to return the solution temperature to 25°C after dissolution of the acid and equilibrium with the ice is reached? The molar heat capacity is 80.8 J/(mol·°C) for the solution, and the molar heat capacity is 75.3 J/(mol·°C) for pure water.
Answer:
15,4g of ice
Explanation:
The dissolution of HNO₃ increases the temperature of the solution thus:
Q = -C×m×ΔT (1)
Where Q is heat produced:
Q = -33 kJ/mol×(0,0130L×14,0M) = -6,00kJ
C is molar heat capacity of solution: 80,8 J/mol°C
m are moles of solution ≈ moles of water = 100mL≡100g×(1mol/18g) = 5,56 moles
And ΔT is final temperature - Initial temperature (X-25°C)
Replacing in (1):
-6000J = -80,8J/mol°C×5,56mol×(X-25°C)
13,4 = X-25°C
X = 38,4°C
Knowing that you want to return the temperature of the system to 25°C, the ice must to absorb 6000J of energy (In the fussion process and increasing each temperature until equilibrium) produced in the dissolution of HNO₃:
6000J = ΔH°fus×X + C×X×ΔT
-Where X are moles of ice-
Replacing:
6000J = 6010J/mol×X + 75,3J/mol°C×X×(38,4°C-25°C)
6000J = 6010J×X + 1006J×X
0,855 = moles of X
In grams:
0,855 moles×(18g/1mol)= 15,4g of ice
I hope it helps!
A galvanic cell consists of a Ni2+/ Ni half-cell and a standard hydrogen electrode.
If the Ni2+/ Ni half-cell standard cell functions as the anode, and the standard cell potential is 0.26 V, what is the standard reduction potential for the Ni2+/Ni half-cell?
A) - 0.26 V
B) - 0.13 V
C) + 0.13 V
D) + 0.26 V
Answer:
A) - 0.26 V
Explanation:
Here Ni undergoes oxidation by loss of electrons, thus act as anode. Standard hydrogen electrode undergoes reduction by gain of electrons and thus act as cathode.
[tex]E^0=E^0_{cathode}- E^0_{anode}[/tex]
Where both [tex]E^0[/tex] are standard reduction potentials.
Given that:- [tex]E^0=0.26\ V[/tex]
Also, [tex]E^0_{[H^{+}/H_2]}=0\ V[/tex]
So,
[tex]E^0=E^0_{[H^{+}/H_2]}-E^0_{[Ni^{2+}/Ni]}[/tex]
[tex]0.26\ V=0\ V- E^0_{[Ni^{2+}/Ni]}[/tex]
[tex]E^0_{[Ni^{2+}/Ni]}=-0.26\ V[/tex]
The standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.
What is standard hydrogen electrode?The standard hydrogen electrode is a reference electrode that was arbitrarily assigned an electrode potential of 0.00V. We know that metals that are above hydrogen in the electrochemical series will always have a negative electrode potential.
As such, the standard reduction potential for the Ni2+/Ni half-cell is - 0.26 V.
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One gram (1.00 g) of vitamin B3 (niacin) is dissolved in water to give 10.0 mL of solution. (a) What is the weight/volume percent concentration of this solution?
Answer:
10 g % w/v
Explanation:
Weight/volume percent concentration means the grams of solute, in 100 mL of solution.
If we have 1 g of B3 in 10mL of solution, the rule of three for the weight volume percent will be:
10mL ___ 1 g
100 mL ____ 10 g
Use the kinetic-molecular theory to explain the following phenomena:a.When 1 mol of a real gas is condensed to a liquid, the volume shrinks by a factor of about 1000.Molecules in a gas are far apart. They are much closer together in a liquid. Molecules in a gas are easily squeezed closer together as the gas is compressed
Answer:
The particles in a gas are comparatively very far from each other. Further an ideal gas has not interactions between the particles i.e repulsion. This explains why a gas that is compressed into a liquid collapses into a volume that is small by a factor of 1000. The particles get much closer to each other to form a liquid and they do not repel each other
Write electron (subshell) configurations for the following atoms: a.15P b. 19K c. 28Ni d. 35Br e. 58Ce
Answer:
15P) 3s² 3p³
19K) 4s¹
28Ni) 3d⁸ 4s²
35Br) 3d¹⁰ 4s² 4p⁵
58Ce) 4f¹ 5d¹ 6s²
Explanation:
1) 15P
Phosphorus has 15 electrons
The order of filling of the energy levels is 1s, 2s, 2p, 3s, 3p, 4s, . . .
The 15 electrons of the phosphorus atom will fill up to the 3p orbital, which will contain three electrons
The electron configuration for 15P is: 1s2 2s2 2p6 3s2 3p3
3s² 3p³
2) 19K
Potassium has 19 electrons
The electron configuration for 19K is: 1s2 2s2 2p6 3s2 3p6 4s1
4s¹
3) 28Ni
Nickel has 28 electrons
The electron configuration for 28Ni is: 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 4 s 2 3 d 8
3d⁸ 4s²
4) 35Br
Bromine has 35 electrons
The electron configuration for 35Br is: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p5
3d¹⁰ 4s² 4p⁵
5) 58Ce
Cerium has 58 electrons
The electron configuration for 58Ce is: 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f1 5d1
4f¹ 5d¹ 6s²
Unknown halogen X2 was added to each of two test tubes which contained aqueous NaCl and aqueous NaI, respectively. When hexane was added to the test tubes, an orange color developed in the top layer of the first test tube, while purple developed in the second.
a. Identify X2.
b. Write a balanced equation for each reaction which occurred.
Answer:
[tex]Br_2[/tex]
Explanation:
The compound [tex]X_2[/tex] is [tex]Br_2[/tex].
The reaction included are:
[tex]Br_2+NaCl--> No\ reaction[/tex] (because bromine is less reactive than chlorine)
But , because of hexane the solution get dilute and its color changes to orange.
Now, NaI is added and we know Br is more reactive than I.
Therefore it replace it.
Reaction: [tex]Br_2+NaI-->NaBr+I_2[/tex]
Purple color develop due to formation of Iodine.
The standard enthalpy of reaction for the dissolution of silica in aqueous HF is 4.6 kJ mol–1 . What is the standard enthalpy of formation of SiF4(g)?
SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1
Compound SiO2(s) HF(aq) H2O(l) SiF4(g)
∆Hºf, kJ mol–1 –910.9 –320.1 –285.8 ???
(A) –1624.3 kJ mol^–1
(B) –1615.1 kJ mol^–1
(C) –949.8 kJ mol^–1
(D) –940.6 kJ mol^–1
Answer:
B) ) –1615.1 kJ mol^–1
Explanation:
since
SiO2(s) + 4 HF(aq) → SiF4(g) + 2 H2O(l) ∆Hºrxn = 4.6 kJ mol–1
the enhalpy of reaction will be
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
where ∆Hºrxn= enthalpy of reaction , ∆Hºfp= standard enthalpy of formation of products , ∆Hºfr = standard enthalpy of formation of reactants , νp=stoichiometric coffficient of products, νr=stoichiometric coffficient of reactants
therefore
∆Hºrxn = ∑νp*∆Hºfp - ∑νr*∆Hºfr
4.6 kJ/mol = [1*∆HºfX + 2*(–285.8 kJ/mol)] - [1*(–910.9kJ/mol) + 4*(–320.1 kJ/mol)]
4.6 kJ/mol =∆HºfX -571.6 kJ/mol + 2191.3 kJ/mol
∆HºfX = 4.6 kJ/mol + 571.6 kJ/mol - 2191.3 kJ/mol = -1615.1 kJ/mol
therefore ∆HºfX (unknown standard enthalpy of formation = standard enthalpy of formation of SiF4(g) ) = -1615.1 kJ/mol
Consider the reaction Mg(s) + I2 (s) → MgI2 (s) Identify the limiting reagent in each of the reaction mixtures below:
a. 100 atoms of Mg and 100 molecules of I2
b. 150 atoms of Mg and 100 molecules of I2
c. 200 atoms of Mg and 300 molecules of I2
d. 0.16 mol Mg and 0.25 mol I2
e. 0.14 mol Mg and 0.14 mol I2
f. 0.12 mol Mg and 0.08 mol I2
g. 6.078 g Mg and 63.455 g I2
h. 1.00 g Mg and 2.00 g I2
i. 1.00 g Mg and 20.00 g I2
Answer:
a) Nor Mg, neither I2 is the limiting reactant.
b) I2 is the limiting reactant
c) Mg is the limiting reactant
d) Mg is the limiting reactant
e) Nor Mg, neither I2 is the limiting reactant.
f) I2 is the limiting reactant
g) Nor Mg, neither I2 is the limiting reactant.
h) I2 is the limiting reactant
i) Mg is the limiting reactant
Explanation:
Step 1: The balanced equation:
Mg(s) + I2(s) → MgI2(s)
For 1 mol of Mg we need 1 mol of I2 to produce 1 mol of MgI2
a. 100 atoms of Mg and 100 molecules of I2
We'll have the following equation:
100 Mg(s) + 100 I2(s) → 100MgI2(s)
This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.
b. 150 atoms of Mg and 100 molecules of I2
We'll have the following equation:
150 Mg(s) + 100 I2(s) → 100 MgI2(s)
I2 is the limiting reactant, and will be completely consumed. There will be consumed 100 Mg atoms. There will remain 50 Mg atoms.
There will be produced 100 MgI2 molecules.
c. 200 atoms of Mg and 300 molecules of I2
We'll have the following equation:
200 Mg(s) + 300 I2(s) →200 MgI2(s)
Mg is the limiting reactant, and will be completely consumed. There will be consumed 200 I2 molecules. There will remain 100 I2 molecules.
There will be produced 200 MgI2 molecules.
d. 0.16 mol Mg and 0.25 mol I2
We'll have the following equation:
Mg(s) + I2(s) → MgI2(s)
Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.16 mol of I2. There will remain 0.09 mol of I2.
There will be produced 0.16 mol of MgI2.
e. 0.14 mol Mg and 0.14 mol I2
We'll have the following equation:
Mg(s) + I2(s) → MgI2(s)
This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.
There will be consumed 0.14 mol of Mg and 0.14 mol of I2. there will be produced 0.14 mol of MgI2
f. 0.12 mol Mg and 0.08 mol I2
We'll have the following equation:
Mg(s) + I2(s) → MgI2(s)
I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.08 moles of Mg. There will remain 0.04 moles of Mg.
There will be produced 0.08 moles of MgI2.
g. 6.078 g Mg and 63.455 g I2
We'll have the following equation:
Mg(s) + I2(s) → MgI2(s)
Number of moles of Mg = 6.078 grams / 24.31 g/mol = 0.250 moles
Number of moles I2 = 63.455 grams/ 253.8 g/mol = 0.250 moles
This is a stoichiometric mixture. Nor Mg, neither I2 is the limiting reactant.
There will be consumed 0.250 mol of Mg and 0.250 mol of I2. there will be produced 0.250 mol of MgI2
h. 1.00 g Mg and 2.00 g I2
We'll have the following equation:
Mg(s) + I2(s) → MgI2(s)
Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles
Number of moles I2 = 2.00 grams/ 253.8 g/mol = 0.00788 moles
I2 is the limiting reactant, and will be completely consumed. There will be consumed 0.00788 moles of Mg. There will remain 0.03322 moles of Mg.
There will be produced 0.00788 moles of MgI2.
i. 1.00 g Mg and 2.00 g I2
We'll have the following equation:
Mg(s) + I2(s) → MgI2(s)
Number of moles of Mg = 1.00 grams / 24.31 g/mol = 0.0411 moles
Number of moles I2 = 20.00 grams/ 253.8 g/mol = 0.0788 moles
Mg is the limiting reactant, and will be completely consumed. There will be consumed 0.0411 moles of Mg. There will remain 0.0377 moles of I2.
There will be produced 0.0411 moles of MgI2.
The limiting reagent in a chemical reaction is the reactant that is completely consumed first. To identify the limiting reagent, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation.
* Reaction mixture a: neither reactant is limiting.
* Reaction mixture b: Mg is the limiting reagent.
* Reaction mixture c: I2 is the limiting reagent.
* Reaction mixture d: Mg is the limiting reagent.
* Reaction mixture e: neither reactant is limiting.
* Reaction mixture f: I2 is the limiting reagent.
* Reaction mixture g: neither reactant is limiting.
* Reaction mixture h: I2 is the limiting reagent.
* Reaction mixture i: Mg is the limiting reagent.
To identify the limiting reagent in a reaction mixture, we need to compare the mole ratios of the reactants to the mole ratio of the products in the balanced chemical equation. The limiting reagent is the reactant that will be completely consumed before the other reactants are consumed.
**Balanced chemical equation:**
Mg(s) + I2 (s) → MgI2 (s)
**Mole ratios:**
Mg : I2 : MgI2 = 1 : 1 : 1
**a. 100 atoms of Mg and 100 molecules of I2**
Since we have the same number of atoms of Mg and molecules of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.
**b. 150 atoms of Mg and 100 molecules of I2**
Since we have more atoms of Mg than molecules of I2, Mg is the limiting reagent. I2 will be completely consumed before Mg is consumed.
**c. 200 atoms of Mg and 300 molecules of I2**
Since we have more molecules of I2 than atoms of Mg, I2 is the limiting reagent. Mg will be completely consumed before I2 is consumed.
**d. 0.16 mol Mg and 0.25 mol I2**
Since we have less moles of Mg than moles of I2, Mg is the limiting reagent. I2 will not be completely consumed.
**e. 0.14 mol Mg and 0.14 mol I2**
Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.
**f. 0.12 mol Mg and 0.08 mol I2**
Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.
**g. 6.078 g Mg and 63.455 g I2**
First, we need to convert the masses of Mg and I2 to moles:
Moles of Mg = 6.078 g / 24.31 g/mol = 0.250 mol
Moles of I2 = 63.455 g / 253.808 g/mol = 0.250 mol
Since we have the same number of moles of Mg and moles of I2, neither reactant is limiting. The reaction will proceed until both reactants are completely consumed.
**h. 1.00 g Mg and 2.00 g I2**
First, we need to convert the masses of Mg and I2 to moles:
Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol
Moles of I2 = 2.00 g / 253.808 g/mol = 0.00791 mol
Since we have less moles of I2 than moles of Mg, I2 is the limiting reagent. Mg will not be completely consumed.
**i. 1.00 g Mg and 20.00 g I2**
First, we need to convert the masses of Mg and I2 to moles:
Moles of Mg = 1.00 g / 24.31 g/mol = 0.0411 mol
Moles of I2 = 20.00 g / 253.808 g/mol = 0.0788 mol
Since we have more moles of I2 than moles of Mg, Mg is the limiting reagent. I2 will not be completely consumed.
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Choose the aqueous solution that has the highest boiling point. These are all solutions of nonvolatile solutes and you should assume ideal van't Hoff factors where applicable. A. 0.100 m K2SO4 B. 0.100 m LiNO3 C. 0.200 m C3H8O3 D. 0.060 m Na3PO4 E. They all have the same boiling point.
Answer: 0.100 m [tex]K_2SO_4[/tex]
Explanation:
Elevation in boiling point is given by:
[tex]\Delta T_b=i\times K_b\times m[/tex]
[tex]\Delta T_b=T_b-T_b^0[/tex] = Elevation in boiling point
i= vant hoff factor
[tex]K_b[/tex] = boiling point constant
m= molality
1. For 0.100 m [tex]K_2SO_{4}[/tex]
[tex]K_2SO_4\rightarrow 2K^{+}+SO_4^{2-}[/tex]
, i= 3 as it is a electrolyte and dissociate to give 3 ions. and concentration of ions will be [tex]3\times 0.100=0.300[/tex]
2. For 0.100 m [tex]LiNO_3[/tex]
[tex]LiNO_3\rightarrow Li^{+}+NO_3^{-}[/tex]
, i= 2 as it is a electrolyte and dissociate to give 2 ions, concentration of ions will be [tex]2\times 0.100=0.200[/tex]
3. For 0.200 m [tex]C_2H_8O_3[/tex]
, i= 1 as it is a non electrolyte and does not dissociate, concentration of ions will be [tex]1\times 0.200=0.200[/tex]
4. For 0.060 m [tex]Na_3PO_4[/tex]
[tex]Na_3PO_4\rightarrow 3Na^{+}+PO_4^{3-}[/tex]
, i= 4 as it is a electrolyte and dissociate to give 4 ions. and concentration of ions will be [tex]4\times 0.060=0.24m[/tex]
Thus as concentration of solute is highest for [tex]K_2SO_4[/tex] , the elevation in boiling point is highest and thus has the highest boiling point.
The aqueous solution of K2SO4 has the highest boiling point among the given options.
Explanation:The boiling point of a solution depends on the concentration and nature of the solute. In this case, we need to compare the boiling points of different aqueous solutions. Higher concentration and the presence of ionic solutes will generally result in a higher boiling point.
Comparing the given options, K2SO4 and Na3PO4 are both ionic compounds, and their solutions will have higher boiling points due to their strong ionic interactions. LiNO3 is also an ionic compound, but its concentration is lower than K2SO4 and Na3PO4. C3H8O3 is a molecular compound and does not dissociate into ions in solution, so its solution will have a lower boiling point than the ionic compounds. Therefore, the aqueous solution of K2SO4 (option A) will have the highest boiling point among the given options.
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What is undergoing oxidation in the redox reaction represented by the following cell notation?
Fe(s) Fe3+(aq) Cl2(g) Cl(aq) Pt
Answer:
[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]
Explanation:
The given cell notation is:-
[tex]Fe_{(s)}|Fe^{3+}_{(aq)}||Cl_2_{(g)}|Cl^-_{(aq)}|Pt[/tex]
Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.
[tex]X\rightarrow X^{n+}+ne^-[/tex]
The oxidation half reaction of the cell is:-
[tex]Fe\rightarrow Fe^{2+}+2e^-[/tex]
1. From the following standard electrochemical potential (and your knowledge of biomedically important metals), answer the questions below. Reaction ΔEº(V) Ti -> Ti+3 2.00 Al -> Al+3 1.70 Cr -> Cr+2 0.56 H2-> 2H+ 0.00 Au -> Au+ -1.68 a. Which metal would be the most anodic? b. Which metal would corrode the most? c. Which metal would be the most reactive? d. Why can Ti and Au be used for making implants, but not Al?
Answer:
The following answer are based only on emf values
a).Most anodic element ,Titanium - Ti
b).Metal That will corrode easily is : Titanium - Ti
c). Most reactive metal is Titanium-Ti
d).Al can't be used for making implants because it undergo corrosion easily and do not form strong passive oxide layer.
Ti forms very strong passive layer which prevents further corrosion.
Au is least reactive and does not corrode.
Note : You have to give answer according to E values
In actual(without considering E values) most anodic, reactive , corrosive element is AlExplanation:
Here the oxidation potential data is given (since elements are getting oxidised)
[tex]\Delta^{0} E[/tex] : Standard Reduction Potential : It is tendency of the element to gain electron and get reduced.
Oxidation Potential is opposite to reduction potential
Standard Oxidation Potential :It is tendency of the element to loose electron and get oxidised.
It is measured in Volts/V
More positive (greater)the value of the oxidation potential , stronger is the tendency of the element to get oxidised.More negative(lesser) the value of oxidation potential lesser the tendency of the element to get oxidised. Hence the element undergo Reduction in comparison to other with more positive value .The main application [tex]\Delta^{0} E[/tex] values is in electrochemical cell to predict the products of the reaction. or predict whether the reaction is spontaneous or not.
Anodic means tendency to undergo oxidationSince [tex]Ti\rightarrow Ti^{+}[/tex] has maximum value of [tex]\Delta^{0}E[/tex] = 2.00V , so it get oxidised as compared to other elements.
Corrode , The metal which oxidise easily will also undergo corrosion easilyFrom the , E values Ti show maximum tendency of oxidation hence it also get corroded easily.
Ti is the most reactive because it oxidised fast as compare to other elements Au (gold) is the least reactive element as its[tex]\Delta^{0}E[/tex] value is negative. So gold don't get corrode and used for making implants. Ti can also be used , since it is most reactive so it react with oxygen present in atmosphere and form a passive layer of oxide which prevent its corrosion. Al also form passive oxide but form very thin layer which is not able to protect it from corrosionFinal answer:
Titanium is the most anodic and most reactive metal given its high standard electrode potential of +2.00V, which also makes it the most prone to corrosion. Titanium and Gold are suitable for biomedical implants due to their corrosion resistance and biocompatibility, while Aluminum is not used due to potential toxicity.
Explanation:
To address the questions related to electrochemical potentials and the reactivity of metals, we must consider both the given standard electrode potentials and practical implications such as corrosion resistance in the case of biomedically important metals.
Most Anodic and Most Corrosive
The most anodic metal would be the one with the lowest reduction potential, which correlates with the highest oxidation potential. Since oxidation is the loss of electrons, the metal with the highest potential to lose electrons (most positive ΔEº) will be the most anodic and the most susceptible to corrosion. In this list, Titanium (Ti) has the highest standard electrode potential at +2.00 V, indicating it would be the most anodic and would corrode the most in an environment where it can react.
Most Reactive Metal
The reactivity of a metal in chemical terms is often reflected by how easily it can lose electrons (oxidize). By the given data, Titanium (Ti) would be the most reactive given its high standard electrode potential for oxidation.
Biomedical Implant Materials
Titanium (Ti) and Gold (Au) are used for biomedical implants because they are biocompatible and highly resistant to corrosion in the body. Titanium forms a stable, protective oxide layer when exposed to air or bodily fluids, preventing further oxidation. While Aluminum (Al) has a high standard reduction potential, which may suggest it is resistive to corrosion, it is not used for implants due to potential toxicity and less stable oxide formation than Ti, which could lead to corrosion in the body.
commercial cold packs consist of solid NH4NO3 and water. In a coffee-cup calorimeter, 5.60g NH4NO3 is dissolved in 100g of water at 22.0C; the temperature falls to 17.9C. Assuming that the specific heat capacity of the solution is 4.18 J/(g*K), calculate the enthalpy of dissolution of NH4NO3, in kJ/mol.
Answer:
-1.37 kJ/mol
Explanation:
The expression for the calculation of the enthalpy of dissolution of [tex[NH_4NO_3[/tex] is shown below as:-
[tex]\Delta H=m\times C\times \Delta T[/tex]
Where,
[tex]\Delta H[/tex] is the enthalpy of dissolution of [tex[NH_4NO_3[/tex]
m is the mass
C is the specific heat capacity
[tex]\Delta T[/tex] is the temperature change
Thus, given that:-
Mass of ammonium nitrate = 5.60 g
Specific heat = 4.18 J/g°C
[tex]\Delta T=17.9-22.0\ ^0C=-4.1\ ^0C[/tex]
So,
[tex]\Delta H=-1.25\times 4.18\times 3.9\ J=-95.9728\ J[/tex]
Negative sign signifies loss of heat.
Also, 1 J = 0.001 kJ
So,
[tex]\Delta H=-0.096\ kJ[/tex]
Also,
Molar mass of [tex[NH_4NO_3[/tex] = 80.043 g/mol
The formula for the calculation of moles is shown below:
[tex]moles = \frac{Mass\ taken}{Molar\ mass}[/tex]
Thus,
[tex]Moles= \frac{5.60\ g}{80.043 \ g/mol}[/tex]
[tex]Moles= 0.06996\ mol[/tex]
Thus, [tex]\Delta H=-\frac{0.096}{0.06996}\ kJ/mol=-1.37\ kJ/mol[/tex]
One of the nuclides in each of the following pairs is radioactive. Predict which is radioactive and which is stable.a. 39/19K and 40/19Kb. 209Bi and 208Bic. nickel-58 and nickel-65
Answer:
a) 39/19 K : stable nuclide, 40/19 K : radioactive nuclide.
b) 209B: stable nuclide, 208Bi : radioactive nuclide
c) nickel-58 : stable nuclide, nickel-65 : radioactive nuclide.
Explanation:
As per the rule, nuclides having odd number of neutrons are generally not stable and therefore, are radioactive.
Mass number (A) = Atomic number (Z) + No. of neutrons (N)
Or, N = A - Z
a)
39/19 K and 40/19 K
Calculate no. of neutrons in 39/19 K as follows:
atomic no. = 19, mass no. 39
N = 39 - 19
= 20 (even no.)
Calculate no. of neutrons in 40/19 K as follows:
atomic no. = 19, mass no. 40
N = 40 - 19
= 21 (odd no.)
Therefore, 39/19 K is a stable nuclide and 40/19 K is a radioactive
nuclide.
b)
209Bi and 208Bi
Calculate no. of neutrons in 209Bi as follows:
atomic no. = 83, mass no. 209
N = 209 - 83
= 126 (even no.)
Calculate no. of neutrons in 208Bi as follows:
atomic no. = 83, mass no. 208
N = 208 - 83
= 125 (odd no.)
Therefore, 209Bi is a stable nuclide and 208Bi is a radioactive nuclide.
c)
nickel-58 and nickel-65
Calculate no. of neutrons in nickel-58 as follows:
atomic no. = 28, mass no. 58
N = 58 - 28
= 30 (even no.)
Calculate no. of neutrons in nickel as follows:
atomic no. = 28, mass no. 65
N = 65 - 28
= 37 (odd no.)
Therefore,nickel-58 is a stable nuclide and nickel-65 is a radioactive nuclide.
To determine which nuclide is radioactive and which is stable, we need to analyze the composition of their nuclei. The number of neutrons in a nuclide plays a crucial role in determining its stability. Based on this, we can predict whether a nuclide is radioactive or stable.
Explanation:In order to predict which nuclide is radioactive and which is stable, we need to determine the composition of their nuclei. Radioactive nuclides have an unstable nucleus that undergoes radioactive decay, while stable nuclides have a more balanced composition.
a. 39/19K has 20 neutrons and is stable, while 40/19K has 21 neutrons and is radioactive.b. 209Bi has 126 neutrons and is stable, while 208Bi has 125 neutrons and is radioactive.c. 58Ni has 30 neutrons and is stable, while 65Ni has 37 neutrons and is radioactive.Therefore, the pairs are:
a. 39/19K (stable) and 40/19K (radioactive)
b. 209Bi (stable) and 208Bi (radioactive)
c. 58Ni (stable) and 65Ni (radioactive)
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Considering factors that affect that solubility which statement below true?
*With a few exceptions, the solubility of most solid solutes in water decreases as the solution temperature increases.
*The solubility of gases in water increases with increasing temperature.
*The solubility of a gas in any solvent is increased as the partial pressure of the gas above the solvent Increases.
*Substances with similar intermolecular attractive forces tend to be insoluble in one another.
Part B
Which of these is the best solvent for nonpolar solutes?
* acetone (CH3COCH3)
*ethanol (CH3CH2OH)
*hexane (C6H14)
*water
Part A:
Therefore the correct statements are as follows.
With a few exceptions, the solubility of most solid solutes in water decreases as the solution temperature increases.(True)
The solubility of gases in water increLases with increasing temperature. (False).
The solubility of a gas in any solvent is increased as the partial pressure of the gas above the solvent increases. (True)
Substances with similar intermolecular attractive forces tend to be insoluble in one another. (False)
solubility.
Part B:
Therefore, hexane (C6H14) is the best solvent for nonpolar solutes.
Part A:
The first statement is true. Generally, the solubility of solid solutes in water tends to decrease with increasing temperature, although there are exceptions.
The second statement is false. The solubility of gases in water typically decreases with increasing temperature. This is described by Henry's law.
The third statement is true. This statement is consistent with Henry's law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid
The fourth statement is false. Substances with similar intermolecular attractive forces tend to be more soluble in one another. Like dissolves like is a general principle in solubility.
Part B:
The best solvent for nonpolar solutes is the one with similar characteristics, meaning it should be nonpolar as well
Acetone is a polar solvent. It is not the best solvent for nonpolar solutes.
Ethanol is a polar solvent. It is not the best solvent for nonpolar solutes.
Hexane is a nonpolar solvent and is the best choice for nonpolar solutes.
Water is a polar solvent. It is not the best solvent for nonpolar solutes.
Therefore, hexane (C6H14) is the best solvent for nonpolar solutes.
From this list, choose all of the ionic compounds. Check all that apply. View Available Hint(s) Check all that apply. POCl3 KOCH2CH3 CH3OH SOCl2 CH3CH2CO2Na NaOH CH2
Answer: [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.
Explanation:
It is known that ionic compounds are the compounds in which one atom transfer its valence electrons to another atom. Hence, during this transfer partial opposite charges tend to develop on the combing atoms due to which strong force of attraction exists between the atoms.
An ionic bond is always formed between a metal and a non-metal.
For example, [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa[/tex], NaOH are all ionic compounds.
On the other hand, a compound formed due to sharing of electrons between the combining atoms is known as a covalent compound. Generally, a non-metal with same or different non-metal tends to form a covalent bond.
For example, [tex]POCl_{3}[/tex], [tex]SOCl_{2}[/tex] etc are all covalent compounds.
Thus, we can conclude that [tex]KOCH_2CH_3[/tex], [tex]CH_3CH_2COONa, NaOH are all ionic compounds.
Methane is also known as natural gas, and its combustion is a very common reaction: CH4 + 2 O2 → CO2 + 2 H2O. Remember that we can use stoichiometry if we know the correct coefficients. So, which of these conversion factors will NOT work for this reaction? NOTE: All numbers located immediately after elemental symbols should be considered subscripts.
A. 1 mole O2 = 32.00 g
B. 2 mole H2O = 18.02 g
C. 2 mole O2 = 1 mole CO2
D. 1 mole CH4 = 2 mole H2O
E. none of the above
Answer:
2 mole H2O = 18.02 g will not work.
Explanation:
Let us check the options one by one:
1 mole of O2 weighs 32 g.
⇒Hence option A is correct.
1 mole of H2O weighs 18 g.
⇒Hence option B is wrong.
From the balanced equation we can say :
1 mole of CH4 reacts with 2 moles of O2 to give 1 mole of CO2 and 2 moles of H2O.
⇒Hence option C and D are correct.
Answer:
Option B: 2 mole H2O = 18.02 g is not correct.
Explanation:
Step 1: The balanced equation
CH4 + 2O2 → CO2 + 2H2O
Step 2: Data given
Molar mass of O2 = 32 g/mol
Molar mass of H2O = 18.02 g/mol
Molar mass of CH4 = 16.04 g/mol
Step 3: Calculate number of mol
For 1 mol of CH4 we need 2 moles of O2 to produce 1 mol of CO2 and 2 moles of H2O
This means option C is correct: For 2 moles O2 consumed, we'll get 1 mol of CO2 produced.
Also Option D is corect: For 1 mole of CH4 we get 2 moles of H2O
Mass = moles * molar mass
Mass of 1 mole O2 = 1 mol * 32.00 g/mol = 32.00 grams
Mass of 2 moles H2O = 2 mol * 18.02 g/mol = 36.04 grams (This means option B is not correct.)
Option B: 2 mole H2O = 18.02 g is not correct.
In a hexagonal-close-pack (hcp) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes is 1:1:2.
What is the general formula for a compound if anions occupy the hcp lattice points and cations occupy all the octahedral holes?What is the general formula for a compound if anions occupy the hcp lattice points and cations occupy all the octahedral and tetrahedral holes?
Explanation:
In a hexagonal-close-pack (HCP) unit cell, the ratio of lattice points to octahedral holes to tetrahedral holes = 1 : 1 : 2
let the :
Number of lattice point = 1x.
Number of octahedral points = 1x
Number of tetrahedral points = 2x
If anions occupy the HCP lattice points and cations occupy half of the octahedral holes.
Number of anions occupying the HCP lattice points, A= 1x
Number of cations occupying the octahedral points, B = 1x
The formula of the compound will be = [tex]A_{1x}B_{1x}=AB[/tex]
If anions occupy the HCP lattice points and cations occupy all of the octahedral and the tetrahedral holes.
Number of anions occupying the HCP lattice points, A= 1x
Number of cations occupying the octahedral points, B = x
Number of cations occupying the tetrahedral points, B = 2x
total number of cations = x + 2x = 3x
The formula of the compound will be = [tex]A_{1x}B_{3x}=AB_3[/tex]
A 0.250 gram chunk of sodium metal is cautiously dropped into a mixture of 50.0 grams of water and 50.0 grams of ice, both at 0 degress celsius. The reaction is:2Na(s)+2H2O(l) yields 2NaOH(aq)+H2(g) Change in H is -368 KJWill the ice melt?
Assuming the final mixture has a specific heat capacity of 4.18 J/g*degrees celsius, calculate the final temperature. The enthalpy of fusion for ice is 6.02 KJ/mol.
Answer:
The ice will not melt, and the temperature will remain at 0°C.
Explanation:
The reaction of the sodium in water is exothermic because heat is being released. In an isolated system, the change in heat must be 0, so the released heat must be absorbed by the ice.
The molar mass of Na is 23 g/mol, so the number of moles that reacted was:
n = 0.250 g/ 23g/mol
n = 0.011 mol
By the reaction:
2 moles ------- -368 kJ
0.011 mol ----- x
By a simple direct three rule:
2x = -4.048
x = -2.024 kJ/mol
So the ice will absorbs 2.024 kJ/mol, which is less than the necessary to melt it (6.02 kJ/mol). Then, the ice will not melt.
The temperature of a pure substance didn't change until all of it has changed of phase, so the temperature must remain at 0°C.
14. 60. g of NaOH is dissolved in enough distilled water to make 300 mL of a stock solution. What volumes of this solution and distilled water, when mixed, will result in a solution that is approximately 1 M NaOH?
The question is incomplete, the complete question is attached below.
Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
Explanation : Given,
Mass of NaOH = 60 g
Volume of stock solution = 300 mL
Molar mass of NaOH = 40 g/mol
First we have to calculate the molarity of stock solution.
[tex]\text{Molarity}=\frac{\text{Mass of }NaOH\times 1000}{\text{Molar mass of }NaOH\times \text{Volume of solution (in mL)}}[/tex]
Now put all the given values in this formula, we get:
[tex]\text{Molarity}=\frac{60g\times 1000}{40g/mole\times 300mL}=5mole/L=5M[/tex]
Now we have to determine the volume of stock solution and distilled water mixed.
Formula used :
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of stock solution.
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted solution.
From data (A) :
[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (B) :
[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 20mL=1M\times V_2\\\\V_2=100mL[/tex]
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (C) :
[tex]M_1=5M\\V_1=60mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From data (D) :
[tex]M_1=5M\\V_1=20mL\\M_2=1M\\V_2=?[/tex]
Putting values in above equation, we get:
[tex]5M\times 60mL=1M\times V_2\\\\V_2=300mL[/tex]
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.
Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
To make a 1 M NaOH solution, you would mix 2 L of the stock solution with 2 L of distilled water.
Explanation:To prepare a 1 M NaOH solution using a stock solution of 60 g NaOH in 300 mL, we need to calculate the amount of stock solution and distilled water needed.
We can start by calculating the number of moles of NaOH in the stock solution:
moles = mass / molar mass => moles = 60 g / (22.99 g/mol + 16.00 g/mol + 1.01 g/mol) = 2 mol NaOH
Since we want a 1 M NaOH solution, we can use the formula: moles = concentration x volume
2 mol = 1 M x volume => volume = 2 mol / 1 M = 2 L
Therefore, we need to mix 2 L of the stock solution with 2 L of distilled water to obtain a 1 M NaOH solution.
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What is the mole fraction of 25.0 g of ethanol (c2h5oh) is added to 75.0 g of water to form 96.4 mL of solution
Answer:
0.116
Explanation:
The molar mass of ethanol is 46.07 g/mol. The moles corresponding to 25.0 g are:
25.0 g × (1 mol/46.07 g) = 0.543 mol
The molar mass of water is 18.02 g/mol. The moles corresponding to 75.0 g are:
75.0 g × (1 mol/18.02 g) = 4.16 mol
The total number of moles is:
total moles = mol ethanol + mol water = 0.543 mol + 4. 16 mol = 4.70 mol
The mole fraction of ethanol is:
X(Ethanol) = mol ethanol / total moles = 0.543 mol / 4.70 mol = 0.116
Carbon Monoxide at 25°C and steam at 150°C are fed to a continuous water-gas shift reactor. The product gas, which contains 50.0 mole% H2, 40.0% ????????2, and the balance H2????, emerges at 500°C at a rate of 2.50m3/h and goes to a condenser. The gas and liquid streams leaving the condenser are in equilibrium at 15°C and 1 bar. The liquid may be taken to be pure w"a) Calculate the % excess steam fed to the reactor and rate of condensation of the water (kg/h) leaving the condensor.b) Calculate the rate (kW) at which heat must be removed from the condensor.c) Calculate the rate of heat transfer (kW) to or from the reactor (state which it is).
The final answer for the solution is Q= -8.8171*[tex]10^{-4}[/tex] kW
Explanation:
[tex]CO +H_2O ⇄ CO_2+H_2[/tex]
basic:
100 moles of product gas
[tex]CO_2[/tex] = 40 moles
[tex]H_2[/tex] = 40 moles
[tex]H_20[/tex] = 20 moles
for 40 moles of[tex]CO_2[/tex] we need 40 moles of CO and 40 moles of [tex]H_2O[/tex]
therefore,
Inlet :
No of moles of CO = 40
No of moles of [tex]H_20[/tex] = 40
but [tex]H_20[/tex] is in the outlet stream = 20 modes
Excess = [tex]\frac{20}{40}\ \times 100[/tex]%
= 50%
50% excess stream is fed to the reactor
a) Rate of conversion:
ρ_water (density ) =[tex]\frac{1g}{cm^{3}}[/tex]
Given volume rate = 3.5 [tex]\frac{cm^{3} }{hr}[/tex]
3.5 * density = 3.5 * [tex]10^{-3}[/tex] [tex]\frac{kg}{hr}[/tex]
b) Rate (kW)
ΔH = ΔH_1 + ΔH_2 +ΔH_3
Rate of condensation Q° =n° ΔH
Q = n° (ΔH_1 + ΔH_2 +ΔH_3)
ΔH_1 is sensible heat
V =[tex]\int\limits^T_S {C_p(T) } \, dT[/tex]
C_p(T) = a + bT + CT^2 +dT^3
H_2O
a=32.218
b=o.192 * 10^-2
c= 1.055 * 10^-5
d= -3.593 * 10^-9
ΔH_2 = ∫(a+bT +a^2 +dT^3)dT (of limits 100 to 500 )
= -13497.6
ΔH_3 =∫(a +bT +T^2 +dT^3)dT (of limit 95 to 100)
= -2751.331
phase change
ΔH_2 = - ΔH_v
ΔH_v = 7.3 @ 100° C
liquid → vapor ⇒ ΔH_v
vapor → liquid ⇒ -ΔH_v
Q= n°(-16324.231)
=350 *[tex]10^{-5}[/tex] (-16324.231)
=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) *[tex]\frac{kg}{hr}[/tex][tex]\frac{J}{mol}[/tex]
=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) * \frac{J}{mol}[/tex]
=[tex]\frac{350* 10^{-5} }{18}[/tex] (-16324.231) *kW
Q= -8.8171*[tex]10^{-4}[/tex] kW
The solution involves various calculations related to a chemical reactor, including determining the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate.
Explanation:This is a complex chemical engineering problem involving numerous calculations, including finding the excess steam fed into the reactor, condensation rate, heat removal rate, and heat transfer rate. Firstly, the percentage excess steam is found by comparing the stochiometric requirement of the reaction with actual steam flow. Next, we use the molar mass of water and the gas constant to find the condensation rate of water. For the heat removal rate, we need to consider the latent heat of condensation along with the energy balance of the condenser.
Lastly, for the heat transfer rate in the reactor, we need to take into account the enthalpy of reaction, heat capacities of reactants and products, and the reactor temperature profile to decide whether heat is added or removed from the reactor.
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You intend to perform the double aldol condensation reaction in this week's experiment using Ketone X and Aldehyde Y. You mix 0.3 moles of Ketone X with 0.4 moles of Aldehyde X . What is your theoretical yield in moles? Round to the nearest tenth
2. The condensation portion of the aldol condensation reaction proceeds via an E1 mechanism
Answer:
1. The theoretical yield is 0.2mols
2. No, aldol condensation reaction does not proceeds via an E1 mechanism.
How many grams of nickel metal are plated out when a constant current of 15.0
A is passed through aqueous NiCl2 for 60.0 minutes?
A) 10.9 gB) 16.4 gC) 32.8 gD) 36.3 g
Answer:
B) 16.4
Explanation:
Given that:-
Current, I = 15.0 A
Time, t = 60.0 minutes
Also, 1 minute = 60 seconds
So, t = [tex]60\times 60[/tex] s = 3600 s
F is Faraday constant = 96485 C
Atomic weight of Nickel = 58.69 g/mol
Also, Charge on Ni in [tex]NiCl_2[/tex] = 2
So, equivalent weight of Ni , E = [tex]\frac{58.69}{2}\ g/mol[/tex] = 29.34 g/mol
Thus, according to the Faraday's Law:-
[tex]W=\frac{EIt}{96485}[/tex]
Where, W is the mass of the metal deposited.
So,
[tex]W=\frac{29.34\times 15\times 3600}{96485}\ g=16.4\ g[/tex]
Weight of Ni metal plated out = 16.4 g
Final answer:
To determine the mass of nickel metal plated out, use Faraday's law of electrolysis and the given values of current and time. The answer is 16.4 grams.
Explanation:
To determine the number of grams of nickel metal plated out, we need to use Faraday's law of electrolysis. The formula is:
Mass (g) = (Current (A) * Time (s)) / (Molar mass (g/mol) * Faraday's constant (C/mol))
Given that the current is 15.0 A and the time is 60.0 minutes (or 3600 seconds), we can use the molar mass of nickel (58.69 g/mol) and Faraday's constant (96,500 C/mol) to calculate the mass:
Mass (g) = (15.0 A * 3600 s) / (58.69 g/mol * 96,500 C/mol) = 16.4 g
Therefore, the answer is Choice B) 16.4 g.
In a saturated solution of Zn(OH)2at 25^oC the value of [OH–] is 2.0 x 10^–6M. What is the value of the solubility-product constant, Ksp, for Zn(OH)2at 25^oC?
(A) 4.0 x 10–18
(B) 8.0 x 10–18
(C) 1.6 x 10–17
(D) 4.0 x 10–12
(E) 2.0 x 10–6
Answer:B
Explanation:
The solubility product is the product of the concentration of all the ions present as found in the rate equation. The solubility product is the equilibrium expression that shows the extent to which a substance dissolves in water. The full solution is shown in the image attached.
Answer:
4.0E^-18
which makes the answer A
Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate anion in the solution. You can assume the volume of the solution doesn't change when the copper(II) acetate is dissolved in it. Be sure your answer has the correct number of significant digits.
Answer:
0.0714 M for the given variables
Explanation:
The question is missing some data, but one of the original questions regarding this problem provides the following data:
Mass of copper(II) acetate: [tex]m_{(AcO)_2Cu} = 0.972 g[/tex]
Volume of the sodium chromate solution: [tex]V_{Na_2CrO_4} = 150.0 mL[/tex]
Molarity of the sodium chromate solution: [tex]c_{Na_2CrO_4} = 0.0400 M[/tex]
Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:
[tex](CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)[/tex]
Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:
[tex]n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol[/tex]
Moles of the sodium chromate solution would be found by multiplying its volume by molarity:
[tex]n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol[/tex]
Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.
Write the net ionic equation for this reaction:
[tex]Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)[/tex]
Notice that acetate is the ion spectator. This means it doesn't react, its moles throughout reaction stay the same. We started with:
[tex]n_{(AcO)_2Cu} = 0.0053515 mol[/tex]
According to stoichiometry, 1 unit of copper(II) acetate has 2 units of acetate, so moles of acetate are equal to:
[tex]n_{AcO^-} = 2\cdot 0.0053515 mol = 0.010703 mol[/tex]
The total volume of this solution doesn't change, so dividing moles of acetate by this volume will yield the molarity of acetate:
[tex]c_{AcO^-} = \frac{0.010703 mol}{0.1500 L} = 0.0714 M[/tex]
The final molarity of the acetate anion in the solution, after copper(II) acetate is dissolved in a sodium chromate solution, is 0.1 M.
Explanation:The acetate anion, CH3 CO₂¯, found in copper(II) acetate, is the conjugate base of acetic acid. When copper(II) acetate is dissolved in a solution of sodium chromate, it yields a solution of inert cations and weak base anions, resulting in a basic solution. Given the initial acetate concentration, if the newly formed acetate ion gives a final acetate concentration of (1.0 × 10−²) + (0.01 × 10−²) = 1.01 × 10-² mol NaCH3 CO₂, this would mean there are both 9.9 × 10-3 mol/L of copper(II) acetate and 1.01 × 10-2 mol/L of sodium chromate.
To compute the molarity of acetate anion, you can divide the moles of the acetate by the volume of the solution, yielding a final molarity of the acetate anion in the solution. Hence the molarity of acetate anion in the solution would be 1.01 x 10^-2 mol /0.101 L = 0.1 M.
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