When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [CO2] = 0.24 M, [H2] = 0.24 M, [H2O] = 0.48 M, and [CO] = 0.48 M. Then an additional 0.34 moles per liter of CO2 and H2 are added. When the reaction comes to equilibrium again at the same temperature, what will be the molar concentration of CO?

Answer:

Kp = 1.37

Explanation:

In order to do this, we need to apply the Hess's law which it states:

"The heat of any reaction  ΔHf°  for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction"

Now, here we don't have data for enthalpy but we do have Kp, so the principle is applied similarly, even is we have Kp.

First thing we should do is to get the overall reaction needed which is the following:

C(s) + CO2(g) <-----> 2CO(g)   (3)

To get to this reaction, we just need to sum the other two reactions, and multiply coefficients if it's needed so:

C(s) + 2H2O(g) <----> CO2(g) + 2H2(g)   Kp1 = 3.79    (1)

H2(g) + CO2(g) <----> H2O(g) + CO(g)    Kp2 = 0.601   (2)

Now, in order to get equation (3), let's look at equations 1 and 2. As we can see, in both equations we have molecules of H2 and water, which aren't present in the overall reaction 3, so we need to get rid of them. In order to do so, if look carefully, you'll see that if you substract molecule of H2 from 1 and from 2, you still have traces of H2 (Because 2H2 - H2 = H2), so, how can we equal both of the molecules?.

That's right, we need to multiply the coefficient of that molecule to equal the coefficient of reaction 1. However keep in mind, that doing so, it will multiply the coefficients of the other molecules too. So doing that we have:

2H2(g) + 2CO2(g) <----> 2H2O(g) + 2CO(g)    Kp = (0.601)²

Now, by multiplying the coefficients of the reaction, it also affects the value of Kp; remember that Kp is a value that you can obtain by doing this:

Kp = P(products) / P(reactants)

If we modify the coefficients by two, Kp is altered:

Kp = P(prod)² / P(react)²

That is why we elevated the value of Kp. Now, summing both equation 2 and 4 we have:

C(s) + 2H2O(g) <----> CO2(g) + 2H2(g)   Kp1 = 3.79

2H2(g) + 2CO2(g) <----> 2H2O(g) + 2CO(g)    Kp = (0.601)²

______________________________________________

C(s) + CO2(g) <-----> 2CO(g)   (3)

Now the value of Kp will be:

Kp = 3.79 x (0.601)² = 1.37

In this exercise we have to calculate the value of the constant, like this:

Kp = 1.37

Using the Hess formula and knowing the equation given as:

[tex]C(s) + CO2(g) \rightarrow 2CO(g)[/tex]

To get to this reaction, we just need to sum the other two reactions, and multiply coefficients if it's needed so:

[tex]C(s) + 2H2O(g) \rightarrow CO2(g) + 2H2(g) \ Kp_1 = 3.79 \\H2(g) + CO2(g) \rightarrow H2O(g) + CO(g) \ Kp_2 = 0.601[/tex]

Multiply the coefficients of the other molecules, we have that:

[tex]2H_2(g) + 2CO_2(g) \rightarrow 2H_2O(g) + 2CO(g) \ Kp = (0.601)^2[/tex]

To find the value of the constant we have to use the formula below and put the values ​​already :

[tex]Kp = P(products) / P(reactants)\\Kp = P(prod)^2 / P(react)^2\\Kp_1 = 3.79\\Kp_2 = (0.601)^2\\Kp = 3.79 * (0.601)^2 = 1.37[/tex]

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Answer:

i think it will be c. The acid in solution A is less concentrated than in solution B and is also a stronger acid than that in solution B. im not for sure

Explanation:

As per the concept of strength and pH of acids the acid in solution A is more concentrated than in solution B as it requires more volume of titrant to reach equivalence point  and is also a weaker acid than that in solution B as it's pH is more than that of solution B.

Acids are defined as substances which on dissociation yield H+ ions , and these substances are sour in taste. Compounds such as  HCl, H₂SO₄ and HNO₃ are acids as they yield H+ ions on dissociation.

According to the number of H+ ions which are generated on dissociation acids are classified as mono-protic , di-protic ,tri-protic and polyprotic  acids  depending on the number of protons which are liberated on dissociation.

Acids are widely used in industries  for  production of fertilizers, detergents  batteries and dyes.They are used in chemical industries for production of chemical compounds like salts which are produced by neutralization reactions.

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Answer:

Explanation:  It has been given that the temperature of the water decreases when chromium chloride is dissolved in water. Thus fall in the temperature explains the fact that the bond energies of the reactants have more energy rather than the products.

a) Thus the heat of the solution is endothermic in nature as more energy is needed to break the reactant molecules.

b) The combined ionic bond strength of CrCl2 and inter molecular forces between water molecules must be stronger than the attractive forces between the  water molecules and chromium and chloride ions as the reaction is endothermic in nature thus more energy would be required to break the bonds between the reactants hence making them more stronger.

Final answer:

The correct degrees of freedom for testing the significance of coefficients in a regression analysis with 17 independent variables and 697 observations is 679. The correct option is c.

Explanation:

The critical value of t for testing the significance of each of the independent variable's coefficients in a regression analysis with 17 independent variables and 697 observations depends on the degrees of freedom (df).

The correct degrees of freedom in this setting is calculated as the number of observations minus the number of independent variables minus one, which is df = 697 - 17 - 1. Therefore, the degrees of freedom (df) for the t-test is 679 (Option c).

Answer:

[tex]C=0.122M[/tex]

Explanation:

Total volume:

[tex]V= 0.5 L + 0.3 L + 0.1 L=0.9 L[/tex]

Total acid (H+) moles:

[tex]n_{acid}=0.5 L * 0.5 mol/L=0.25 mol[/tex]

Total base (OH-) moles:

[tex]n_{base}=0.3 L*0.2 mol/L + 0.1 L*0.4mol/L*2=0.14mol[/tex]

Excess acid:

[tex]n_{ex}=0.25 mol - 0.14mol=0.11 mol[/tex]

Concentration:

[tex]C=\frac{0.11mol}{0.9L}=0.122M[/tex]

Answer:

The correct answer is option d. a polyatomic anion.

Explanation:

Hello! Let's solve this!

The ending "ate" or "ite" indicates that there is a negative polyatomic ion (polyatomic anion). This means that the termination of the name will indicate the valence number of the element, if the number used is the highest or the lowest.

An example is:

CO-3: is the carbonate ion

SO-4: is the sulfate ion

We conclude that the correct answer is option d. a polyatomic anion.

An ate or ite  suffix at the end of a compound name usually indicates that the compound is a polyatomic anions.

Polyatomic ion:

The ions that contain more than one type of atom. Most of polyatomic ions are anions that are named with suffix -ate or -ite.

For examples-

[tex]\bold {CO_3^-^2}[/tex]- Carbonate ion, end with -ate.

[tex]\bold {SO_3^-^2}[/tex] Sulfite ion, end with -ite.

Therefore, an ate or ite  suffix at the end of a compound name usually indicates that the compound is a polyatomic anions.

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Answer:

True

Explanation:

All the above statements buttress the fact that the larger molecule, the greater the magnitude of London forces between the molecules. Each of the statements above is a confirmation/explanation of this general rule.

The statement is true; larger molecules possess stronger London forces due to their larger polarizable electron clouds, making instantaneous dipoles more likely and the resulting dispersion forces stronger.

The statement that the larger the molecules of a substance, the stronger the London forces between them is true. This occurs because a larger molecule possesses more electrons, increasing the chances of having its electron cloud distorted, thereby facilitating the formation of instantaneous dipoles. Larger molecules have more polarizable electron clouds, which means they can be distorted more easily and thus, induce stronger London dispersion forces.

London forces are a type of van der Waals force and exist in all molecules, whether polar or nonpolar. They arise from the movements of electrons that create temporary dipoles, which can then induce dipoles in adjacent atoms or molecules. This interaction leads to dispersion forces that increase in strength as the size of the molecules increases. Therefore, larger molecules tend to have higher boiling and melting points due to the stronger London forces present.

Answer: To speed up the reaction

Explanation:

Potassium iodide is used in elephant toothpaste reaction for the decomposition of hydrogen peroxide by removal of oxygen from the solution ,so that reaction gets completed. Potassium iodide is the catalyst and we know that the concentration of catalyst will not change throughout the reaction so, Potassium iodide will not be consumed in the foam making process.

Answer:

The value of n is 14.

Explanation:

To calculate the concentration of solute, we use the equation for osmotic pressure, which is:

[tex]\pi=icRT[/tex]

where,

[tex]\pi[/tex] = osmotic pressure of the solution = 57.1 Torr =[tex]\frac{57.1}{760} atm = 0.07513 atm[/tex]

1 atm = 760 Torr

i = Van't hoff factor = 2 (electrolytes)

c = concentration of solute = ?

R = Gas constant = [tex]0.0820\text{ L atm }mol^{-1}K^{-1}[/tex]

T = temperature of the solution = [tex]25^oC=[273.15+25]=298.15 K[/tex]

Putting values in above equation, we get:

[tex]c=\frac{\pi}{iRT}=\frac{0.07513 atm}{2\times 0.0821 atm L/mol K\times 298.15 K}[/tex]

[tex]c=0.001535 mol/L[/tex]

Assuming that molality and molarity in such a dilute solution.

c = m (Molality)

The salt is soluble in water to the extent of 0.036 g per 100 g of water at 25°C

[tex]Molaity=\frac{\text{Mass of solute}}{\text{molar mass of solute(M)}\times \text{Mass of solvent in kg}}[/tex]

Molality of the solution = m = 0.001535 mol/L

[tex]\frac{0.036 g}{M\times 0.1 kg}=0.001535 mol/kg[/tex]

M = 234.53 g/mol

Molar mass of [tex]LiC_nH_{2n+1}O_2[/tex] : M

M = [tex]7 g/mol\times 1 + 12 g/mol \times n +1 g/mol\times (2n+1)+2\times 16 g/mol[/tex]

[tex]234.53 g/mol=7 g/mol\times 1 + 12 g/mol \times n +1 g/mol\times (2n+1)+2\times 16 g/mol[/tex]

n = 14

The value of n is 14.

Final answer:

To find the value of n in LiCnH2n+1O2, the osmotic pressure and solubility data were used, assuming complete dissociation. However, the van 't Hoff factor may be less than the ideal value due to high lattice energy of lithium compounds, leading to potential incomplete dissociation. Further experiments or calculations are necessary to accurately determine n.

Explanation:

To determine an appropriate value of n in the formula LiCnH2n+1O2 for the lithium salt used in lubricating grease, we will use the osmotic pressure measurement given and the assumption of complete dissociation. Given that the osmotic pressure is 57.1 torr, which converts to 0.075 atm (since 1 atm = 760 torr), we can use the formula for osmotic pressure Π = iMRT, where Π is the osmotic pressure in atmospheres, i is the van 't Hoff factor, M is the molarity, R is the ideal gas constant (0.0821 L⋅atm/mol⋅K), and T is the temperature in Kelvin (298 K, assuming 25°C room temperature). In this case, the salt fully dissociates into Li+ and the organic anion, hence the van 't Hoff factor, i, should be 2.

First, we determine the molarity of the solution. We have a solubility of 0.036 g per 100 g of water, which equates to 0.036 g in 0.1 kg of water. Assuming an approximate molar mass for the lithium salt to be around 6 (Li) + 14n (for the CnH2n+1 part) + 32 (for O2) = 38 + 14n, we can use the relation:

Π = (2)(M)(0.0821)(298)

0.075 atm = (2)(M)(0.0821)(298)

By isolating M, we find M = 0.075 / (2 * 0.0821 * 298) = 0.001538 mol/kg. Given the weight of the salt used, we can calculate the molar mass: 0.036g / 0.001538 mol/kg = 23.41 g/mol. Using the approximate molar mass 38 + 14n = 23.41, we can solve for n:

38 + 14n = 23.41

14n = -14.59

n ≈ -1.04

However, since n cannot be negative and must be an integer for an organic molecule, an error must have occured in our calculations or assumptions. Lithium compounds do have high lattice energies, and hence it's possible that the ionic compound doesn't fully dissociate.

Considering the information provided about Lithium compounds, lattice energy, and real solutions being less than ideal, the van 't Hoff factor i for the lithium salt may actually be less than the ideal value of 2, indicating incomplete dissociation. Additional experiments or refined calculations would be required to determine the actual value of n.

Answer:

The mass of KIO3 is 42.8 grams

Explanation:

Step 1: Data given

Volume = 2.00 L

Molarity = 0.100 molar

Molar  mass of KIO3 = 214 g/mol

Step 2: Calculate moles

Moles = Molarity * volume

Moles = 0.100 M * 2.00 L

Moles = 0.200 moles

Step 3: Calculate mass of KIO3

Mass KIO3 = moles KIO3 * molar mass KIO3

Mass KIO3 = 0.200 moles * 214 g/mol

Mass KIO3 = 42.8 grams

The mass of KIO3 is 42.8 grams

To prepare a 2.00-liter, 0.100-molar solution of KIO₃, weigh out 42.8 grams of KIO3, dissolve it in water, and transfer it to a 2.00-liter volumetric flask, filling to the mark with water. Proper labeling and documentation are crucial for accurate and safe experimentation.

To prepare 2.00 liters of a 0.100-molar solution of KIO₃ (potassium iodate) with a molecular weight of 214 g/mol, you need to calculate the mass of KIO₃ required and then follow these steps:

Calculate the moles of KIO₃ needed:

Moles = Molarity × Volume (in liters)

Moles = 0.100 mol/L × 2.00 L = 0.200 moles

Calculate the mass of KIO3 required:

Mass (g) = Moles × Molecular Weight

Mass (g) = 0.200 moles × 214 g/mol = 42.8 grams

Weigh out 42.8 grams of KIO₃ accurately using a laboratory balance.

Place the weighed KIO₃ sample into a suitable container (like a beaker).

Add distilled or deionized water to the container and dissolve the KIO₃ completely. Stir the solution to ensure thorough mixing.

Transfer the solution to a 2.00-liter volumetric flask if available, and then add more water to reach the 2.00-liter mark, making sure the solution is well-mixed.

Cap and label the volumetric flask with the appropriate details, including the chemical name, molarity, and date of preparation.

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Final answer:

Joe correctly prepared the 1.0 M potassium phosphate solution by adding water up to the calibration mark after dissolving the solute, whereas Jennifer's method would result in a concentration less than 1.0 M.

Explanation:

To prepare a 1.0 M aqueous solution of potassium phosphate properly, the student should weigh out the necessary amount of solute and then add it to a volumetric flask that is already partially filled with water. After the solute dissolves, the water should be added to the calibration mark to ensure the correct final volume of the solution. In the scenario described, Joe correctly prepared the solution because he added water to the calibration line after dissolving the potassium phosphate. If Jennifer filled the flask to the calibration mark before adding the solute, her solution would have a slightly greater volume than 1.0 liter, which would result in a concentration of less than 1.0 M. It’s crucial to follow these steps to ensure the solution’s concentration matches the intended molarity.

Answer:

Substance B is not involved in the rate-determining step of the mechanism, but is involved in subsequent steps

Explanation:

According to the law of mass action:-

The rate of the reaction is directly proportional to the active concentration of the reactant which each are raised to the experimentally determined coefficients which are known as orders. The rate is determined by the slowest step in the reaction mechanics.

Order of in the mass action law is the coefficient which is raised to the active concentration of the reactants. It is experimentally determined and can be zero, positive negative or fractional.

The order of the whole reaction is the sum of the order of each reactant which is raised to its power in the rate law.  

Thus,  Given that:- The change in the concentration of B does not affect the rate of the reaction. Hence, the order of B must be zero.

Hence, the answer is;- substance B is not involved in the rate-determining step of the mechanism, but is involved in subsequent steps

The reaction is zero-order concerning B, which explains why doubling its concentration doesn't affect the reaction rate. However, the order concerning A cannot be determined from the provided information.

For the reaction 2A(g) + B(g) → 2C(g), when the concentration of substance B is doubled and the rate of the reaction remains unchanged, it suggests that the reaction rate is independent of the concentration of B. This means that the reaction is zero-order with respect to B; the concentration of B does not affect the reaction rate. The rate law for this reaction could therefore be expressed as r = k[A]²[B], indicating that the reaction is second-order with respect to A and zero-order with respect to B.

Answer:

I¹³¹ , Ir¹⁹² ,Sm¹⁵³

Explanation:

Iodine 131 has Symbol I¹³¹

Iridum-192 has Symbol Ir¹⁹²

Samarium-153 has Symbol Sm¹⁵³

Answer:

Lavoisier made an important contribution to chemistry by the law of conservation of mass

Explanation:

The law of conservation of mass tell us, that the mass doens't change in a system. You have the same mass at the begining and in the end of a reaction.

Matter is neither created, nor destroyed in a chemical reaction.

This law also states that mass of reactants is the same of products in any chemical reaction

Answer:

Identifying substances by weight

Answer:

A

Explanation:

Hybridization is simply a phenomena which involves the mixing of orbitals to form new ones. It is simply a way of forming a whole new set of orbitals from old ones.

In XeF4, the type of hybridization that we have is the sp3d2 hybridization. This simply means we have one s orbital, mixing with 3 p orbitals and 2 d orbitals. These orbitals mix together to form the new hybrid orbital. It must be noted that the hybridization takes place in the central atom xenon Xe. The valence shell of xenon contains 2 electrons in the 5s orbital and 6 electrons in the 5p orbitals. In the state of excitement, 2 of the electrons in the outermost 5p orbitals get excited and promoted to the 5d orbitals. This causes a total of four unpaired electrons in which the four chlorine atom can attach with.

In ammonia, there are three hydrogen atoms which seek to join forces with a single nitrogen atom. It must be known that there are 8 electrons around the central atom nitrogen. There are a set of lone pair which are non bonding while the other three are in connection with the 3 hydrogen atoms. Instead of the molecule having 1s and 3p orbitals, they show hybridization to give sp3 hybrid orbital

Answer:

E =  3.6 x 10⁶ J/mol

f =   1.1 x 10 ¹⁶ s⁻¹

λ =  2.6 x 10⁻⁸ m

Explanation:

Rydberg´s equation for hydrogen-like atoms is:

1/λ = Z²Rh (1/n₁² - 1/n₂²)

where  λ = wavelength  

Z² = atomic number of hydrogen-like atom  

 Rh= Rydberg´s constatn                                       

n₁ = principal quantum number of initial state                        

 n₂ = principal quantum number of final state

We also know that E = h(c/ λ ) = hf, where f is frequency equal to c/λ,  so we have all the information needed to answer the questions.

a)  We are asked the energy to remove the electron from 1 mol of B⁴⁺ , that means the transition is from  n₁ = 3 to n₂ = ∞. The term 1/n₂ approaches zero in the infinity so:

Working in  SI units

1/λ =  5² x1.097 x 10⁷ m⁻¹ ( 1/3² - 0) = 3.0 x 10⁷ m⁻¹

E= h(c/ λ )= hc(1/ λ) = 6.626 x 10⁻³⁴ J/s x  3 x 10⁸ m/s x  (3.0 x 10⁷ m⁻¹)

= 6.0x 10⁻¹⁸ J

This is the energy per atom, so per mol of atoms  is

= 6.0x 10⁻¹⁸ J/atom x 6.022 x 10²³ atoms/mol = 3.6 x 10⁶ J/mol

b) f and λ from a transition n= 3 to n=2

1/ λ = 5² x1.097 x 10⁷ m⁻¹ x ( 1/2² - 1/3²) = 3.8 x 10⁷  m⁻¹    ⇒

λ = 1/ 3.8 x 10⁷  m⁻¹ = 2.6 x 10⁻⁸ m

f = 3 x 10⁸ m/s / 2.6 x 10⁻⁸ m = 1.1x 10 ¹⁶ s⁻¹

Answer:

At STP, 760mmHg or 1 atm and OK or 273 degrees celcius

Explanation:

The standard temperature and pressure is the temperature and pressure at which we have the molecules of a gas behaving as an ideal gas. At this temperature and pressure, it is expected that the gas exhibits some properties that make it behave like an ideal gas.

This temperature and pressure conform some certain properties on a gas molecule which make us say it is behaving like an ideal gas. Ordinarily at other temperatures and pressures, these properties are not obtainable

Take for instance, one mole of a gas at stp occupies a volume of 22.4L. This particular volume is not obtainable at other temperatures and pressures but at this particular temperature and pressure. One mole of a gas will occupy this said volume no matter its molar mass and constituent elements. This is because at this temperature and pressure, the gas is expected to behave like an ideal gas and thus exhibit the characteristics which are expected of an ideal gas

The given question is incomplete. The complete question is as follows.

At which temperature and pressure will a sample of neon gas behave most like an ideal gas?

Choices are as follow:

(1) 100 K and 0.25 atm

(2) 100 K and 25 atm

(3) 400 K and 0.25 atm

(4) 400 K and 25 atm

Explanation:

At low pressure and high temperature there exists no force of attraction or repulsion between the molecules of a gas. Hence, gases behave ideally at these conditions.

Whereas at low temperature there occurs a decrease in kinetic energy of gas molecules and high pressure causes the molecules to come closer to each other.

As a result, there exists force of attraction between the molecules at low temperature and high pressure and under these conditions gases are known as real gases.

For the given conditions, 400 K and 0.25 atm  depicts low pressure and high temperature.

Thus, we can conclude that at 400 K and 0.25 atm  a sample of neon gas behave most like an ideal gas.

Explanation:

For what I can see, is missing the concentration of [Ag+] in the half-cell. To calculate it:

Niquel half-cell

Oxidation reaction: [tex]Ni \longrightarrow Ni^{2+}+2 e^-[/tex]

[tex]E=E^0 - \frac{R*T}{n*F}*ln(1/[Ni^{2+}])[/tex]

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

[tex]E=-0.23V - \frac{8.314*298}{2*96500}*ln(1/0.014M)[/tex]

[tex]E=-0.285V[/tex]

Silver half-cell

Reduction reaction: [tex]Ag^+ + e^- \longrightarrow Ag[/tex]

[tex]E=E^0 - \frac{R*T}{n*F}*ln(1/[Ag+])[/tex]

[tex]E_{cell}=E_{red} - E_{ox}[/tex]

[tex]E_{red}=1.12 V + (-0.855V)=0.835V[/tex]

Assuming T=298 K / R=8.314 J/mol K / F=96500 C

[tex]0.835V=0.8V - \frac{8.314*298}{1*96500}*ln(1/[Ag+])[/tex]

[tex][Ag+]=0.26 M[/tex]

Final answer:

A concentration cell is an electrochemical cell in which the anode and cathode compartments have different concentrations of a reactant. The initial cell voltage of +1.12 V indicates that the reaction is spontaneous and will proceed in the forward direction.

Explanation:

A concentration cell is an electrochemical cell in which the anode and cathode compartments are identical except for the concentration of a reactant. In this case, the concentration of Ni2+(aq) is different in the two compartments of the voltaic cell. As the reaction proceeds and the concentrations equilibrate, the measured potential difference between the two compartments will decrease until it reaches zero. The initial cell voltage of +1.12 V indicates that the reaction is spontaneous and will proceed in the forward direction.

Answer:

In organic chemistry, the structural formula shows the bonding and general layout of the molecule.

Explanation:

It can also help in naming the molecule, as many compounds with the same molecular formula have different structural formulas, for example cycloalkanes and alkenes, or aldehydes and ketones.

It tells us about the constituents of the compound, or in other words, the functional groups present. This enables us to predict what kind of properties the compound has and what kind of reactions it can undergo.

It can also help us determine the stereochemistry (shape and spatial orientation) of the compound. This is especially important in organic chemistry and organic chemstry, since certain important reactions will proceed if and only if a molecule with the right shape is employed.

Answer:

Final concentration of Mg2+ = 0.20 M

Explanation:

Concentration of [tex]MgCl_2(aq)\ (M_1)[/tex] = 0.60 M

Volume of [tex]MgCl_2(aq)\ (V_1)[/tex] = 200 mL

Volume of distilled water added = 400 mL

Final volume of the soution = 200 mL + 400 mL

                                              = 600 mL

Final concentration of solution = [tex]M_2[/tex]

The final concentration is calculated as follows:

[tex]M_1 V_1=M_2V_2\\0.60 \times 200= M_2 \times 600\\M_2=\frac{0.60\times 200}{600} =0.20\ M[/tex]

Therefore, final concentration of the solution is 0.20 M.

[tex]MgCl_2(aq)[/tex] exists in the solution as Mg2+ and 2Cl-.

Therefore, concentration of Mg2+ is same as the final concentration of solution.

Final concentration of Mg2+ = 0.20 M

The concentration of magnesium ion, Mg²⁺ in the resulting solution is 0.2 M.

To solve this question, we'll begin by calculating the molarity of the diluted solution. This can be obtained as follow:

Volume of stock solution (V₁) = 200 mL

Molarity of stock solution (M₁) = 0.60 M

Volume of diluted solution (V₂) = 200 + 400 = 600 mL

The molarity of the diluted solution can be obtained as follow:

M₁V₁ = M₂V₂

0.6 × 200 = M₂ × 600

120 = M₂ × 600

Divide both side by 600

M₂ = 120 / 600

Thus, the molarity of the diluted solution of MgCl₂ is 0.2 M

Finally, we shall determine the concentration of Mg²⁺ in the diluted solution. This is illustrated below:

MgCl₂(aq) —> Mg²⁺(aq) + 2Cl¯(aq)

From the balanced equation above,

1 mole MgCl₂ dissolves to produce 1 mole Mg²⁺.

Therefore,

0.2 M MgCl₂ will also produce 0.2 M Mg²⁺.

Thus, the concentration of Mg²⁺ in the resulting solution is 0.2 M.

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Answer:88.8%

Explanation:

%yield= actual yield/theoretical yield * 100/1

41.9/47.2*100/1= 88.8%

Answer: concentration

Explanation:

Concentration refers to the amount of a substance present in a sample. The more molecules of a substance present in a sample, the greater its concentration. The less molecules of a substance in a sample, the lesser the concentration. We are often concerned about analytically determining the concentration of a substance using diverse analytical methods in chemistry.

Explanation:

Solubility is determined by the principle , "like dissolves like" .

i.e. , if a compound is polar then it will dissolve in a polar compound only , and

if a compound is non - polar then it will dissolve in a non - polar compound only .

Hence , from the question ,

Water is a polar molecule , and hence it will dissolve only the polar molecule , i.e. , from the given options the polar molecule is , iii. K₂SO₄

Hexane , is a non - polar molecules ,  hence it will dissolve only the non polar molecule , i.e. , from the given options the non polar molecule is i. Butane .

The substance soluble in water is [tex]\rm ii. CH_3COOH\;\;\; iii. K_2SO_4[/tex]

The substance soluble in hexane is  i. Butane

Solubility is the measure that how much solute will be soluble in liquid.

The substance soluble in water are polar solvents. Only these substances are soluble in water.

The substance that are not soluble in water are called non-polar solvents.

The acetic acid  and potassium sulfate is soluble in water because it has polar ends that dissolve with the water.

The butane is soluble in hexane not in water because it has non-polar ends.

Thus, The substance soluble in water is [tex]\rm ii. CH_3COOH\;\;\; iii. K_2SO_4[/tex]

The substance soluble in hexane is  i. Butane.

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Answer:

1) Cu(NO₃)₂(aq) + 2NaOH (aq) -> cu(OH)₂(s)+ 2NaNO₃ (aq)

   reactants ratio = Cu(NO₃)2 : NaOH  = 1:2

2) FeSO₄ (aq) + 2NaOH (aq) ->Fe(OH)₂ (s) + Na₂SO₄ (aq)

  reactants ratio = FeSO₄  : NaOH  = 1:2

3) Fe(NO₃)₃ (aq) + 3NaOH (aq) -> Fe(OH)₃ (s) + 3NaNO₃ (aq)

  reactants ratio = Fe(NO₃)₃  : NaOH  = 1:3

Explanation:

For any reaction to occur completely, the ratio of the reactants should be the ratio of their coefficients is the balanced chemical equation.

A balanced equation is an equation for a chemical reaction in which the number of atoms for each element in the reaction and the total charge is the same for both the reactants and the products.

In our case, we get greatest amount of precipitate only when the reactants completely undergo chemical change and turn into products.

so, when reactants are taken in the ration of balanced equation, the reactants completely form products, where here one of the products is precipitate.

1) Cu(NO₃)₂(aq) + 2NaOH (aq) -> cu(OH)₂(s)+ 2NaNO₃ (aq)

   reactants ratio = Cu(NO₃)2(aq) : NaOH (aq) = 1:2

2) FeSO₄ (aq) + 2NaOH (aq) ->Fe(OH)₂ (s) + Na₂SO₄ (aq)

  reactants ratio = FeSO₄ (aq) : NaOH (aq) = 1:2

3) Fe(NO₃)₃ (aq) + 3NaOH (aq) -> Fe(OH)₃ (s) + 3NaNO₃ (aq)

  reactants ratio = Fe(NO₃)₃ (aq) : NaOH (aq) = 1:3  

Final answer:

To determine the ratio of reactants that would result in the greatest amount of precipitate in each reaction, understanding the stoichiometry of the balanced equations is crucial.

Explanation:

For each reaction, the ratio of reactants that would result in the formation of the greatest amount of precipitate can be determined by examining the stoichiometry of the balanced equation.

In the reaction Fe(NO3)3(aq) + 3NaOH(aq) -> Fe(OH)3(s) + 3NaNO3(aq), a 1:3 ratio of Fe(NO3)3 to 3NaOH would produce the most precipitate because each mole of Fe(NO3)3 reacts with 3 moles of NaOH to form Fe(OH)3.

By understanding the stoichiometry of the reactions, you can determine the optimal ratio of reactants to maximize the formation of precipitate.

Answer:

The correct statements are A amine groups are fully protonated and can be described with the chemical formula NH3+ B carboxylic acid functional groups are de protonated and can be be described with the chemical formula COO-  C Amine functional groups are positively.

Explanation:

If we study the biochemical structure of an amino acid we wil see that an amino group or -NH2 is present at one end and a carboxylic group or COOH is present at another end.

  Now the fact the that when an amino acid exist as zwitterion it contain same number of positive charge as well as same number of negative charge.So during zwitterion formation the carboxylic acid or -COOH liberates a proton and exist as COO- whereas the amine group accepts that proton and exist as NH3+.

  Beside this the amine group -NH2 after the formation of zwitterion gains a positive charge and exist as -NH3+.

Answer : The number of moles of [tex]NO_2[/tex] gas added must be, 1.59 moles.

Explanation :

Equilibrium constant : It is defined as the equilibrium constant. It is defined as the ratio of concentration of products to the concentration of reactants.

The equilibrium expression for the reaction is determined by multiplying the concentrations of products and divided by the concentrations of the reactants and each concentration is raised to the power that is equal to the coefficient in the balanced reaction.

The given equilibrium reaction is,

                        [tex]SO_2(g)+NO_2(g)\rightleftharpoons SO_3(g)+NO(g)[/tex]

Initial conc.    2.86               x             0          0

At eqm.       (2.86-1.30)    (x-1.30)     1.30     1.30

The expression of [tex]K_{eq}[/tex] will be,

[tex]K_{eq}=\frac{[SO_3][NO]}{[SO_2]NO_2]}[/tex]

Now put all the given values in this expression, we get:

[tex]3.70=\frac{(1.30)\times (1.30)}{(2.86-1.30)\times (x-1.30)}[/tex]

[tex]x=1.59[/tex]

Thus, the number of moles of [tex]NO_2[/tex] gas added must be, 1.59 moles.

Answer:

He will decide which drink is to be served to whom, by the use of litmus paper.

Explanation:

The litmus paper is the most common indicator to determine the acidity or basicity of a solution. Blue litmus paper changes its color to red when a solution changes from basic to acidic while red litmus paper changes its color to blue when the opposite occurs (acid → basic).

First of all the litmus paper strip, pH indicator, is immersed in a solution and allowed to pass between 10 and 15 seconds while keeping the strip submerged.  Afterwards it is removed, and then the strip compares the color. If the color is diffuse, there is a color scale where it is determined which solution has alkaline or acidic pH

Answer:

thermosphere

Explanation:

Answer:

The vapor pressure of water at 50 °C  is 93.7 torr.

Explanation:

The expression for Clausius-Clapeyron Equation is shown below as:

[tex]\ln P = \dfrac{-\Delta{H_{vap}}}{RT} + c [/tex]

Where,  

P is the vapor pressure

ΔHvap  is the Enthalpy of Vaporization

R is the gas constant (8.314×10⁻³ kJ /mol K)

c is the constant.

For two situations and phases, the equation becomes:

[tex]\ln \left( \dfrac{P_1}{P_2} \right) = \dfrac{\Delta H_{vap}}{R} \left( \dfrac{1}{T_2}- \dfrac{1}{T_1} \right)[/tex]

Given:

[tex]P_1[/tex] = 23.8 torr

[tex]P_2[/tex] = ?

[tex]T_1[/tex] = 25°C

[tex]T_2[/tex] = 50 °C  

ΔHvap = 43.9 kJ/mol

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (25 + 273.15) K = 298.15 K  

T = (50 + 273.15) K = 323.15 K  

[tex]T_1[/tex] = 298.15 K  

[tex]T_2[/tex] = 323.15 K

So,  applying in the above equation as:-

[tex]\ln \:\left(\:\frac{23.8}{P_2}\right)\:=\:\frac{43.9}{8.314\times 10^{-3}}\:\left(\:\frac{1}{323.15}-\:\frac{1}{298.15}\:\right)[/tex]

[tex]\frac{23.8}{P_2}=e^{\frac{43.9}{8.314\times \:10^{-3}}\left(\frac{1}{323.15}-\frac{1}{298.15}\right)}[/tex]

[tex]23.8=\frac{1}{e^{\frac{1097500}{801030.39216}}}P_2[/tex]

[tex]P_2=23.8e^{\frac{1097500}{801030.39216}}=93.7\ torr[/tex]

The vapor pressure of water at 50 °C  is 93.7 torr.

To calculate the vapor pressure of water at a different temperature, you can use the Clausius-Clapeyron equation, which relates the vapor pressure of a liquid to its temperature. Input the known values into the equation, and solve for the vapor pressure at the new temperature.

The key to answering this question is the Clausius-Clapeyron relationship, an equation used in Physical Chemistry to describe the relationship between the vapor pressure and temperature of a liquid. The equation is { ln(P2/P1) = -ΔHvap/R (1/T2 - 1/T1) }. Where P1, T1 represent the initial condition, in this case, the vapor pressure at 25 degrees Celsius (converted to Kelvin - 298.15 K) and seed pressure; P2 and T2 represent the final conditions, P2 being the one we want to calculate and T2 the final temperature (50 degrees Celsius converted to Kelvin - 323.15 K); ΔHvap is the enthalpy of vaporization, and R is the ideal gas constant (8.314 J/mol*K). Solve the equation for P2 to find the final vapor pressure under the new conditions.

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Answer: Aerobic respiration.

Aerobic respiration is the process that takes place in the presence of oxygen and produces the most ATP.

ETC, or Electron Transport chain specifically produces the most ATP.

Oxidative phosphorylation is the process that takes place in the presence of oxygen, producing almost 20 times more ATP than glycolysis alone. It occurs in the mitochondria or inner part of the cell membrane and is a part of the aerobic catabolism of glucose, which involves glycolysis, the Krebs cycle, and the electron transport chain.

The process that takes place in the presence of oxygen and produces nearly 20 times as much ATP as glycolysis alone is known as oxidative phosphorylation. This process involves the movement of electrons through a series of reactions to a final electron acceptor, which is oxygen. These reactions take place in the mitochondria of eukaryotic cells or the inner part of the cell membrane of prokaryotic cells.

Most of the ATP generated during the aerobic catabolism of glucose is actually from oxidative phosphorylation. This process involves a series of complex steps, beginning with glycolysis, followed by the Krebs cycle (also known as the citric acid cycle or tricarboxylic acid cycle), and ending with the electron transport chain and chemiosmosis.

During glycolysis, glucose is broken down to form pyruvate. In the presence of oxygen, pyruvate continues on to the Krebs cycle, where more ATP is generated along with the energy-carrying molecules NADH and FADH2. It's these molecules that pass on to the next step— the electron transport chain— where oxidative phosphorylation occurs. The result is the production of a large amount of ATP from the energy of the electrons removed from hydrogen atoms in the glucose molecule.

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