Answer:
Moles of potassium chlorate = 0.02976 moles
Explanation:
At standard pressure and temperature,
22.4 L of a gas consists of 1 mole
Thus, given, volume of [tex]O_2[/tex] = 1.0 L
So,
1 L of a gas consists of [tex]\frac{1}{22.4}[/tex] mole
Moles of oxygen gas = 0.04464 moles
The reaction is shown below as:-
[tex]2KClO_3\rightarrow 2KCl+3O_2[/tex]
3 moles of oxygen gas are produced when 2 moles of potassium chlorate undergoes reaction.
So,
1 mole of oxygen gas are produced when [tex]\frac{2}{3}[/tex] moles of potassium chlorate undergoes reaction.
Thus,
0.04464 mole of oxygen gas are produced when [tex]\frac{2}{3}\times 0.04464[/tex] moles of potassium chlorate undergoes reaction.
Moles of potassium chlorate = 0.02976 moles
From the decomposition reaction 2KClO₃(s) → 2KCl(s) + 3O₂(g), the number of moles of KClO₃ to be decomposed to generate 1.0 L of O₂ gas at standard temperature and pressure (STP) is 0.030.
The balanced chemical reaction for the decomposition of potassium chlorate (KClO₃) is the following:
2KClO₃(s) → 2KCl(s) + 3O₂(g) (1)
We can find the number of moles of O₂ gas with the Ideal gas equation:
[tex] PV = nRT [/tex]
Where:
P: is the pressure = 1.0 atm (at STP conditions)
V: is the volume = 1.0 L
R: is the gas constant = 0.082 L*atm/(K*mol)
T: is the temperature = 273 K (at STP conditions)
n: is the number of moles =?
The number of moles of O₂ gas is:
[tex] n_{O_{2}} = \frac{PV}{RT} = \frac{1.0 atm*1.0 L}{0.082 L*atm/(K*mol)*273 K} = 0.045 \:moles [/tex]
From reaction (1), we have that 2 moles of KClO₃ produce 3 moles of O₂, so the number of moles of KClO₃ resulting from the decomposition is:
[tex] n_{KClO_{3}} = \frac{2\:moles\:KClO_{3}}{3\:moles\:O_{2}}*0.045\:moles\:O_{2} = 0.030 \:moles [/tex]
Therefore, the number of KClO₃ moles to be decomposed is 0.030.
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According to the law of multiple proportions, when water forms, the mass ratio of hydrogen to oxygen is variable.
Answer & Explanation:
The law of multiple proportion states that two elements A and B can react with each other to form multiple compounds while one of the two elements remain fixed.
For Example
Hydrogen reacts with oxygen to form:
Water (H2O)
Hydrogen peroxide (H2O2)
The Law of Multiple Proportions states that the mass ratio of elements in different compounds is a simple whole number ratio. In the case of water, the mass ratio of hydrogen to oxygen is always 1:8.
Explanation:The Law of Multiple Proportions is a principle in chemistry that states that when two elements combine to form different compounds, the ratio of their masses will be a simple whole number ratio. In the case of water, the ratio of hydrogen to oxygen by mass is always 2:16 or 1:8. This means that for every 1 gram of hydrogen in water, there are 8 grams of oxygen. The mass ratio of hydrogen to oxygen in water is therefore constant, not variable.
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The larger the molecules of a substance, the the London forces between them. A larger molecule has more electrons and a greater of having its electron cloud distorted from its nonpolar shape. Thus instantaneous dipoles are more likely to form in larger molecules. The electron clouds in larger molecules are also larger, so the average distance between the nuclei and the electrons is greater; as a result, the electrons are held and shift more easily to create a dipole.
True / False.
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.
Explanation: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.
A student wishes to prepare 2.00 liters of 0.100–molar KIO3 (molecular weight 214). The proper procedure is to weigh out.
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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For the reaction 2A(g) + B(g) → 2C(g), when the concentration of substance B in the reaction above is doubled, all other factors being held constant, it is found that the rate of the reaction remains unchanged. The most probable explanation for this observation is that
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.
As the number of covalent bonds between two atoms increases the distance between the atoms,the distance between the atoms_______ and the strength of the bond between them_______.
a) increases,increases.
b) is unprecdictable.
c) increases,decreases.
d) decreases,decreases.
e) decreases,increases.
Answer:
E
Explanation:
As the number of covalent bonds increase, atoms are drawn closer together hence the distance between the atoms decreases. However, the strength of the bonds increases. The bond energy of nitrogen gas (N2) is high because of the strong triple bond between its atoms hence the gas is inert.
As the number of covalent bonds between two atoms increases, the distance between the atoms decreases and the strength of the bond between them increases.
Explanation:As the number of covalent bonds between two atoms increases, the distance between the atoms decreases and the strength of the bond between them increases. Covalent bonds are formed when atoms share electrons. The stronger the bond, the shorter the distance between the atoms.
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ClO2- + HCOOH(aq) ---- HClO2 (aq) + HCOO-(aq) Keq<1 what are the relative strengths of the acids and bases in the reaction represented by the equation above ?Acid stength Base strengthA. HClO2< HCOOH ClO2-< HCOO-B. HClO2> HCOOH ClO2< HCOO-
Answer:
B
Explanation:
The species on the right are the more stable acids and bases. A weak acid always has a strong conjugate base as in HCOO- and HCOOH. The acid is a weak acid because it has a strong conjugate base. Similarly, a strong acid has a weak conjugate base as in ClO2- and HClO2. ClO2- is a weak base hence it quickly abstracts a proton to form the acid.
HClO₂ is a stronger acid than HCOOH, and ClO₂⁻ is a weaker base than HCOO⁻. Therefore, the correct answer is B.
To determine the relative strengths of acids and bases in the given reaction:
ClO₂⁻ + HCOOH → HClO₂ + HCOO⁻
We use the Brønsted-Lowry definition of acids and bases. According to the reaction and knowing that equilibrium constants (Keq) of less than one (<1) favor reactants:
Acid strength: Since the reaction lies more towards the reactants (HCOOH and ClO₂⁻), this implies that HClO₂ is a stronger acid than HCOOH.Base strength: Similarly, ClO₂⁻ is a weaker base than HCOO⁻.Thus, the correct answer is: B. HClO₂ > HCOOH, ClO₂⁻ < HCOO⁻
What volume of chlorine gas at 45.3oC is needed to react with 14.2g of sodium to form NaCl at 1.72atm?
Answer:
9.4L
Explanation:
In this particular question, we will need to write a balanced chemical equation for the reaction between sodium and chlorine to form sodium chloride.
Na + Cl ——> NaCl
Hence, we can see that 1 mole of chlorine reacts with 1 mole of sodium.
Now, we need to find the exact number of moles of chlorine atom that reacted with 14.2g of sodium. To do this, we simply divide the mass of the sodium by the atomic mass of the sodium which is 23.
Hence, the mass of sodium reacted is 14.2/23 which equals 0.617 moles
Simply because we have the mole ratio to be 1 to 1, it can be deduced that the number of moles of sodium reacted is also 0.617moles
Now, to get the volume of chlorine, we can use the ideal gas equation.
This is :
PV = nRT
V = nRT/P
Given:
v = ?
n = number of moles = 0.617 moles in this case
T = temperature = 45.3 + 273.15 = 318.45K
P = Pressure = 1.72 atm = 1.72 * 101325 pa = 174,279 Pa
R = molar gas constant = 8314.462L.Pa/K.mol
Inserting all these into the equation will yield:
V = (0.617 * 8314.462 * 318.45)/174,279
V = 9.4L
__________ is the process where chemical ions are transported along the intact skin by an electrical current.
Answer:
Ionotophoresis
Explanation:
Ionotophoresis is the process when "chemical ions are transported through intact skin using electrical current"and is usually used to treat skin infections or in order to create other benefits to the skin.
In order to use this method we need to use a low voltage current direct and we need to ensure a continuous mode with a long pulse duration in order to the ions can flow through the surface of interest.
So then the answer would be:
Ionotophoresis is the process where chemical ions are transported along the intact skin by an electrical current.
Many hospitals use radioisotopes for diagnosis and treatment or in palliative care. Three radioisotopes used in medicine are given. Write the isotope symbol for each radioisotope. Replace the question marks with the proper integers. Replace the letter X with the proper element symbol.
a) Iodine-131:b) Iridium-192:c) Samarium-153:
Answer:
I¹³¹ , Ir¹⁹² ,Sm¹⁵³
Explanation:
Iodine 131 has Symbol I¹³¹
Iridum-192 has Symbol Ir¹⁹²
Samarium-153 has Symbol Sm¹⁵³
A body cools from 60°C to 50°C in 10 minutes. Find its temperature(in °C) at the end of next 10 minutes
Answer:
40 degrees Celsius
Explanation:
In this kind of question, we need to know what is called a temperature gradient. This means we need to know how the temperature changes with time.
To calculate this , we subtract the initial temperature from the final temperature so that we get the total temperature change. Afterwards, we can now divide the total temperature change by the time taken so we get the gradient.
The total temperature change is 10 degrees celcius. The gradient is 10 degrees Celsius divided by 10 minutes which is equal to 1 degree Celsius per minute. Hence, we conclude that the temperature drops at a rate of 1 degree Celsius per minute.
At the end of next ten minutes, we have a temperature drop of another 10 degrees Celsius. If we subtract this from the final temperature I.e 50, this means we now have a new final temperature of 40 degrees Celsius
3. A Beer's law plot for Cu2+ was experimentally obtained. The slope of the Beer's law plot was 435 L/mol (with a y-intercept = 0.001). A Cu2+ solution of unknown concentration had an absorbance of 0.75. What is the molar concentration of Cu2+ in the unknown solution?
Beer's law states that the absorbance of a solution is directly proportional to its concentration. Given the molar absorptivity and absorbance for a Cu2+ solution, we can calculate the concentration to be approximately 0.00172 M.
Explanation:The subject of this question is Beer's Law in the field of chemistry. Beer's Law states that the absorbance of a solution is directly proportional to its concentration, and mathematically it can be expressed as A = εcl. In this formula, A is the absorbance of the solution, ε is the molar absorptivity (the slope in the Beer's law plot), c is the concentration of the solution, and l is the length of the light path through the solution.
In your case, you have been given the slope (ε) from the Beer's law plot, which is 435 L/mol, and the absorbance (A) of the Cu2+ solution, which is 0.75. You need to find the concentration (c), so you can rearrange the formula to c = A / ε.
Substituting the given values into this equation gives: c = 0.75 / 435 = 0.0017241379310344827586206896551724 M
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Language of chemistry
Answer:
A
Explanation:
Pico(p) has a value of 10^-12.
Nano(n) has a value of 10^-9.
A. 374ps= 374×10^-12 s
B. 3.74ps= 374×10^-14 s
C. 374ns= 374×10^-9 s
At which temperature and pressure will a sample of neon gas behave most like an ideal gas?
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.
Aqueous solutions of sodium sulfide and copper(II) chloride are mixed together. Which statement is correct?
A. CuS will precipitate from solution.
B. No reaction will occur.
C. NaCl will precipitate from solution.
D. A gas is released.
E. Both NaCl and CuS precipitate from solution.
Answer:
Correct statement is A.
Explanation:
Let's see the reactions:
2Na⁺ + S⁻² → Na₂S (aq)
This salt is soluble
CuCl₂ (aq) → Cu²⁺ + 2Cl⁻
This is an insoluble salt.
Cu²⁺ (aq) + 2Cl⁻ (aq) + 2Na⁺ (aq) + S⁻² (aq) → 2NaCl (aq) + CuS (s) ↓
No molecules or gas are formed. (Option D, FALSE)
NaCl does not precipitate, because it'a soluble salt (OPTION E or C are false)
Option B is also false. There is a reaction, of precipitation.
Answer:
A.
Explanation:
Firstly, let’s write a chemical equation for the observation:
Na2S + CuCl2 → CuS + 2NaCl
It should be noted that the copper sulphide is a solid and the sodium chloride is in the aqueous form.
B. Is incorrect. A chemical reaction will occur. The chemical reaction will yield copper sulphide and aqueous sodium chloride.
C. is incorrect. Sodium chloride is in the solution form and cannot precipitate in that form.
D. Is incorrect as no gas is released
E. Is incorrect, only copper sulphide will precipitate.
_____ is formed when fossil fuels or turpentine are incompletely burned in the presence of sunlight.
Answer:
Smog
Explanation:
Smog is formed when fossil fuels or turpentine are incompletely burned in the presence of sunlight.
Smog is type of highly dirty air that contain various impurity in it. A form of visible air pollution consists of oxides of ammonia, sulfur , asbestos, haze, and other particulate matter. Man-made smog is derived from emissions from coal burning, vehicle emissions, industrial emissions, forest and agricultural fires, as well as photochemical reactions from these emissions.
Answer:
smog
Explanation:
Consider the exothermic reaction
2C2H6(g)+7O2(g)→4CO2(g)+6H2O(g)
Calculate the standard heat of reaction, or ΔH∘rxn, for this reaction using the given data. Also consider that the standard enthalpy of the formation of elements in their pure form is considered to be zero.
Reactant or product
ΔH∘f (kJ/mol) C2H6(g) -84.7 CO2(g) -393.5 H2O(g) -241.8
Express your answer to four significant figures and include the appropriate units.
Answer:
The answer to your question is -2855 J
Explanation:
Reaction
2C₂H₆ + 7O₂ ⇒ 4CO₂ + 6H₂O
Formula
Heat of reaction = ΔHrxn = ΣΔHrxn products - ΣΔHrxn reactants
Substitution
ΔHrxn = { 4(-393.5) + 6(-241.8)} - {2(-84.7) + 7(0)}
ΔHrxn = {-1574 -1450.8} - {-169.4}
ΔHrxn = -3024.8 + 169.4
ΔHrxn = -2855.4 J
The standard heat of reaction, or ΔH∘rxn, for the given exothermic reaction is calculated using Hess's Law and the provided standard enthalpies of formation, ΔH∘f. By subtracting the sum of the standard enthalpies of formation of the reactants from that of the products, we find ΔH∘rxn is -2850.0 kJ/mol.
Explanation:To calculate the standard heat of reaction, or ΔH∘rxn, for the given exothermic reaction, we utilize the principle of Hess's Law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for each step of the reaction. So, we can use the standard enthalpies of formation (ΔH∘f) provided for each reactant and product.
Specifically, ΔH∘rxn is calculated by subtracting the sum of the standard enthalpies of formation of the reactants from the sum of the standard enthalpies of formation of the products, each multiplied by their stoichiometric coefficients. For this reaction, this gives:
ΔH∘rxn = [4(-393.5) + 6(-241.8)] - [2(-84.7) + 7(0)] = -2850.0 kJ/mol
So, the standard heat of reaction for this exothermic reaction is -2850.0 kJ/mol.
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Two monoprotic acid solutions (A and B) are titrated with identical NaOH solutions. The volume to reach the equivalence point for solution A is twice the volume required to reach the equivalence point for solution B, and the pH at the equivalence point of solution A is higher than the pH at the equivalence point for solution B.
a. The acid in solution A is less concentrated than in solution B and is also a weaker acid than that in solution B.
b. The acid in solution A is more concentrated than in solution B and is also a stronger acid than that in solution B.
c. The acid in solution A is less concentrated than in solution B and is also a stronger acid than that in solution B.
d. The acid in solution A is more concentrated than in solution B and is also a weaker acid than that in solution B.
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.
What is an acid?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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Determine the oxidation number for the indicated element in each of the following compounds: (a) Co in LiCoO2, (b) Al in NaAlH4, (c) C in CH3OH (methanol), (d) N in GaN, (e) Cl in HClO2, (f) Cr in BaCrO4.
The oxidation numbers for the elements Co, Al, C, N, Cl, and Cr in the given compounds are +3, -3, -2, -3, +1, and +6 respectively.
Explanation:The oxidation numbers for the elements in the compounds are determined as follows:
(a) In LiCoO2, Co has an oxidation number of +3, as lithium contributes +1 and each of the two oxygen atoms contribute -2.
(b) In NaAlH4, Al has an oxidation number of -3, as sodium contributes +1, while hydrogen as a metal hydride contributes -1 each adding up to -4.
(c) In CH3OH, C has an oxidation number of -2, as hydrogen contributes +1 each from the three hydrogen atoms and the fourth hydrogen attached to oxygen contributes -1 and oxygen contributes -2.
(d) In GaN, N has an oxidation number of -3, as gallium contributes +3.
(e) In HClO2, Cl has an oxidation number of +1, as hydrogen contributes +1, and each of the two oxygen atoms contribute -2.
(f) In BaCrO4, Cr has an oxidation number of +6, as barium contributes +2 and each of the four oxygen atoms contribute -2.
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The oxidation numbers for the given elements in their compounds are as follows: Co in LiCoO₂ is +3, Al in NaAlH₄ is +3, C in CH₃OH is -2, N in GaN is -3, Cl in HClO₂ is +3, and Cr in BaCrO₄ is +6.
(a) Co in LiCoO₂: The oxidation number of Co is +3. This is because Lithium (Li) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule LiCoO₂ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Co, we get: 1(+1) + x + 2(-2) = 0, which simplifies to x = +3.
(b) Al in NaAlH₄: The oxidation number of Al is +3. This is because Sodium (Na) has an oxidation number of +1, and Hydrogen (H) has an oxidation number of -1. Since the molecule NaAlH₄ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Al, we get: 1(+1) + x + 4(-1) = 0, which simplifies to x = +3.
(c) C in CH₃OH (methanol): The oxidation number of C is -2. This is because Hydrogen (H) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule CH₃OH is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of C, we get: 4(+1) + x + 1(-2) = 0, which simplifies to x = -2.
(d) N in GaN: The oxidation number of N is -3. This is because Gallium (Ga) has an oxidation number of +3. Since the molecule GaN is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of N, we get: 1(+3) + x = 0, which simplifies to x = -3.
(e) Cl in HClO₂: The oxidation number of Cl is +3. This is because Hydrogen (H) has an oxidation number of +1, and Oxygen (O) has an oxidation number of -2. Since the molecule HClO₂ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Cl, we get: 1(+1) + x + 2(-2) = 0, which simplifies to x = +3.
(f) Cr in BaCrO₄: The oxidation number of Cr is +6. This is because Barium (Ba) has an oxidation number of +2, and Oxygen (O) has an oxidation number of -2. Since the molecule BaCrO₄ is neutral, the sum of oxidation numbers is zero. Letting 'x' be the oxidation number of Cr, we get: 1(+2) + x + 4(-2) = 0, which simplifies to x = +6.
What is the standard enthalphy change ΔHo, for the reaction represented above? (ΔHof of C2H2(g) is 230 kJ mol-1; (ΔHof of C6H6(g) is 83 kJ mol-1;)
Answer:
-608KJ/mol
Explanation:
3 C2H2(g) -> C6H6(g)
ΔHrxn = ΔHproduct - ΔHreactant
ΔHrxn= ΔHC6H6 - 3ΔHC2H2
ΔHrxn = 83 - 3(230)
ΔHrxn = -608
I don’t understand how to find these please help
Answer:
8 shared electrons
Explanation:
When you are looking for the number of shared electrons via the equation:
S = N-A
Where:
S = means the shared electrons
N = Needed electrons
A = available electrons
"Needed electrons" means how many electrons does it need to have a noble gas configuration, in this case, to complete the octet rule.
"Available electrons" means how many valence electrons is actually available considering the compound or the elements involved in the compound.
To get the needed electrons, treat the elements involved separately. We have a silicon (Si) atom and 4 chlorine (chlorine) atoms in this compound. Let's list it down first:
Number of atoms
Si 1
Cl 4
Next step is to determine how many electrons should it have in its outer shell to achieve the octet rule. Both of them in this case would be 8. Multiply that by the number of atoms and add up the needed electrons to determine how many you will need for this particular compound.
Number of atoms Electrons to achieve Octet Needed
Si 1 x 8 = 8
Cl 4 x 8 = 32
TOTAL: 40
This is now our N. N = 40 electrons
Next step is to determine how many we actually have. Your clue in determining how many valence electrons the atom has is the group. Silicon is in Group 4A, this means it has 4 valence electrons. Chlorine is in Group7A, so ths means it has 7 valence electrons.
So first we write the number of atoms again, then in the next column, you write down the actual number of valence electrons and multiply them. Sum it up to see how many electrons available in this particular compound.
Number of atoms Valence electrons Available
Si 1 x 4 = 4
Cl 4 x 7 = 28
TOTAL: 32
This is now our A. A = 32 electrons
Now we apply this:
S = N - A
N = 40 electrons
A = 32 electrons
S = 40 - 32 = 8
Number of shared electrons is 8
The vapor pressure of water and the partial pressure of hydrogen contribute to the total pressure of 715 torr. What is the partial pressure of just H2(g) in atmospheres?
Answer:
0.91 atm is the partial pressure of just hydrogen gas.
Explanation:
Vapor pressure of water , p= 0.0313 atm
Partial pressure of hydrogen gas = [tex]p_{H_2}[/tex]
Total pressure of the water vapors and hydrogen gas = P = 715 Torr
1 atm = 715 Torr
[tex]715 Torr=\frac{715}{760} atm=0.94 atm[/tex]
According Dalton's law of partial pressure:
[tex]P=p+p_{H_2}[/tex]
[tex]0.94 atm=0.0313 atm+p_{H_2}[/tex]
[tex]p_{H_2}=0.94 atm - 0.0313 atm =0.9087 atm \approx 0.91 atm[/tex]
0.91 atm is the partial pressure of just hydrogen gas.
What is the molarity of a nano3 solution if 25.0 ml of a 0.200 m nano3 solution is diluted?
samples of matter that contain only one type of element or compound. However, mixtures—samples of matter containing two or more substances physically combined—are more commonly encountered in nature than are pure substances. Similar to a pure substance, the relative composition of a mixture plays an important role in determining its properties. The relative amount of oxygen in a planet’s atmosphere determines its ability to sustain aerobic life. The relative amounts of iron, carbon, nickel, and other elements in steel (a mixture known as an “alloy”) determine its physical strength and resistance to corrosion. The relative amount of the active ingredient in a medicine determines its effectiveness in achieving the desired pharmacological effect. The relative amount of sugar in a beverage determines its sweetness. In this section, we will describe one of the most common ways in which the relative compositions of mixtures may be quantified.
SLO Demonstrate use of dimensional analysis converting from one unit to another through the mole Directions: Answer the following questions. Set-up all problems using the factor-label method of dimensional analysis and show all your work and units. 1. What volume would be occupied by 9.45 x 10²⁴ molecules of CO₂ gas at STP? 2. How many calcium atoms would be in a 100 gram sample of calcium metal? How many grams are in 5.6 x 10²³ atoms of Zinc? 3. Calculate the number of molecules in 4.56-g of Pb(NO₃)₂.
Answer:
1. 351.62L
2. 15.05 * 10^23 molecules
3. 54.6g
4. 8.29 * 10^21 molecules
Explanation:
1. First we will need to calculate the number of moles present in such number of molecules. We know that one mole contains 6.02 * 10^23 molecules. Hence to calculate the number of moles in 9.45 * 10^24 carbon iv oxide Miley, we simply divide. And that is (9.45 * 10^24) / ( 6.02 * 10^23) = 15.7 moles
At s.t.p, a mole of a gas will occupy a volume of 22.L. Hence 15.7 moles will occupy a volume of 15.7 * 22.4 = 351.62L
2. We first need to calculate the number of moles of calcium metal present in 100g. To do this we simply divide the mass by the atomic mass of calcium. The atomic mass of calcium is 40 amu. The number of moles is thus 100/40 = 2.5 moles
One mole contains 6.02 * 10^23 number of molecules. Hence 2.5 moles will contain 2.5 * 6.02 * 10^23 = 15.05 * 10^23 molecules
3. Firstly we need to calculate the number of moles contained in those number of molecules. What we simply need do is to divide by 6.02 * 10^23
Since the power of 10 is same, we simply divide 5.06 by 6.02 = 0.84 moles
To get the mass, we simply multiply by the atomic mass of zinc. The atomic mass of zinc is 65a.m.u
The mass is thus 0.84 * 65 = 54.6g
4. Firstly we will need to calculate the number of moles here. To do this we divide the mass by the molar mass. The molar mass of lead nitrate is 331g/mol
The number of moles is thus 4.56/331 = 0.014 moles
1 mole contains 6.02 * 10^23 molecules, then 0.014 moles will contain 0.014 * 6.02 * 10^23 molecules = 8.29 * 10^21 molecules
Which equation is correctly balanced? Group of answer choices LaTeX: A) Ca\:+Cl_2\:\longrightarrow\:CaCl C a + C l 2 ⟶ C a C l LaTeX B) Ca\:+Cl_2\:\longrightarrow\:Ca_2Cl C a + C l 2 ⟶ C a 2 C l LaTeX C) 2H_2\:+O_2\:\longrightarrow\:2H_2O 2 H 2 + O 2 ⟶ 2 H 2 O LaTeX D) H_2\:+O_2\:\longrightarrow\:H_2O
Answer: [tex]2H_2+O_2\longrightarrow 2H_2O[/tex]
Explanation:
Law of conservation of matter : It states that matter can neither be created nor be destroyed but it can only be transformed from one form to another form.
Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.
A) [tex]Ca+Cl_2\longrightarrow CaCl[/tex]
B) [tex]Ca+Cl_2\longrightarrow Ca_2Cl[/tex]
C) [tex]2H_2+O_2\longrightarrow 2H_2O[/tex] is correctly balanced as the number of atoms of hydrogen and oxygen are equal on the both side of the reaction.
D) [tex]H_2+O_2\longrightarrow H_2O[/tex]
A deficiency in B6 (pyridoxal phosphate) would negatively impact which of the following pathways: 1. Metabolism of homocysteine to cysteine 2. Metabolism of phenylalanine to tyrosine 3. Conversion of methyl malonyl-CoA to succinyl-CoA
Answer:
The correct answer is 1.
Explanation:
The metabolism of homocysteine produces a sulfur amino acid that is normally formed from methionine during the fulfillment of its function as a donor of methyl groups. Metabolic fate such as remethylation and transsulfuration, involving the enzymatic forms of the vitamins folacin, B12, and B6, gives rise to homocysteine and mixed disulfides including so-called protein-linked homocysteine, the main form circulating in plasma. B6 deficiency would have a direct impact on the metabolism of homocysteine to cysteine.
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A 10.0-gram sample of liquid water at 23.0°C absorbs 209 joules of heat.
What is the final temperature of the liquid water sample?
a) 5.0°C b) 28.0°C c) 18.0°C d) 50.0°C
Answer:
The final temperature of the water is 28.0 °C
Explanation:
Step 1: Data given
Mass of liquid water = 10.0 grams
Temperature = 23.0 °C
Heat absorbed = 209 Joules
Since heat was absorbed by the water, you must have a positive value for
Δ T
Step 2: Calculate final temperature
q = m*c* ΔT
⇒ with m = the mass of the water = 10.0 grams
⇒ with c = the specific heat of water = 4.184 J/g°C
⇒ with ΔT = The change in temperature = T2 - T1 = T2 - 23.0 °C
⇒ with q = the heat absorbed = 209 Joule
209 = 10.0 * 4.184 * ΔT
ΔT = 5
ΔT = 5 = T2 - 23
T2 = 28 °C
The final temperature of the water is 28.0 °C
The final temperature of the liquid water sample is 28.0°C.
Explanation:The specific heat capacity of water is 4.184 J/(g°C). To find the final temperature of the water sample, we can use the equation:
q = mcΔT
Where q is the heat absorbed, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
Given that m = 10.0 grams, ΔT = final temperature - 23.0°C, and q = 209 joules, we can rearrange the equation to solve for the final temperature:
209 joules = (10.0 grams)(4.184 J/(g°C))(final temperature - 23.0°C)
Simplifying the equation gives:
final temperature - 23.0°C = 209 joules / (10.0 grams)(4.184 J/(g°C))
final temperature - 23.0°C = 4.9886854°C
final temperature = 28.0°C
The dentist and at least one other dental auxiliary must be present, in the treatment room, during the administration of nitrous oxide
Answer: The dentist and at least one other dental auxiliary must be present, in the treatment room, during the administration of nitrous oxide. True.
Explanation: there is a minimum standard of care while performing nitrous oxide inhalation by a medical personnel and in addition shall maintain under continuous direct supervision auxiliary personnel who shall be capable of reasonably assisting in procedures, problems and emergencies incident to the used of nitrous oxide inhalation sedation.
Is a Magnesium ribbon (Mg) a pure substance or a mixture ?
Answer:
mixture
Explanation:
Answer: Yes or True
Explanation:
Blood is a Mixture. It's a Homo Mixture because it's made up of more than one substance. a shiny magnesium ribbon is burned in air, to form a grayish powder called magnesium oxide. is this oxide an element, compound or mixture. ... It's a compound, because oxygen and magnesium make magnesium oxide.
Use the Bohr model to calculate the radius and the energy of the B⁴⁺ ion in the n 3 state. How much energy would be required to remove the electrons from 1 mol of B⁴⁺ in this state? What frequency and wavelength of light would be emitted in a transition from the n 3 to the n 2 state of this ion? Express all results in SI units.
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
Copper(II) chloride and lead(II) nitrate react in aqueous solutions by double replacement. Write the balanced chemical equation, the overall ionic equation, and the net ionic equation for this reaction. If 11.95 g of copper(II) chloride react, what is the maximum amount of precipitate that could be formed?
Answer:
Balanced chemical equation:
CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)
Overall ionic equation:
Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)
Net ionic equation:
2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)
Amount of precipitate:
24.72 g
Explanation:
First, let's identify the compounds and the products of the reaction. Copper(II) chloride is CuCl₂, and lead(II) nitrate is Pb(NO₃)₂, after the reaction the products will be PbCl₂ and Cu(NO₃)₂, the first one is an insoluble salt, which will precipitate, and the second one is a soluble salt. So, the balanced chemical equation will be:
CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)
The ionic equation is done by putting the ions that are presented when the compound ionization at the aqueous solution. The metals have their charged expressed in the name of the compound, and chloride and nitrate have charge -1:
Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)
The net ionic equation is the simplified ionic equation. So let's eliminate the ions that are presented on both sides of the equation:
2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)
The stoichiometry of the reaction is 1 mol of CuCl₂ for 1 mol of PbCl₂(the precipitate). The molar mass of the compounds are:
CuCl₂ = 134.452 g/mol
PbCl₂ = 278.106 g/mol
1 mol of CuCl₂ ------------ 1 mol of PbCl₂
Transforming in mass:
134.452 g of CuCl₂ ----------------- 278.106 g of PbCl₂
11.95 g of CuCl₂ ---------------- x
By a simple direct three rule:
134.452x = 3323.3667
x = 24.72 g of PbCl₂
Balanced chemical equation:
CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)
Overall ionic equation:
Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)
Net ionic equation:
2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)
The maximum amount of precipitate that could be formed is 24.72 g.
Balanced chemical reaction:Copper(II) chloride is CuCl₂, and lead(II) nitrate is Pb(NO₃)₂, these react together with to form PbCl₂ and Cu(NO₃)₂. The balanced chemical equation will be:
CuCl₂(aq) + Pb(NO₃)₂(aq) → Cu(NO₃)₂(aq) + PbCl₂(s)
The metals have their charged expressed in the name of the compound, and chloride and nitrate have charge -1. The ionic equation will be:
Cu²⁺(aq) + 2Cl⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → Cu²⁺(aq) + 2NO₃⁻(aq) + PbCl₂(s)
Net-ionic equation:
2Cl⁻(aq) + Pb²⁺(aq) → PbCl₂(s)
The stoichiometry of the reaction is 1 mol of CuCl₂ for 1 mol of PbCl₂(the precipitate). The molar mass of the compounds are:
CuCl₂ = 134.452 g/mol
PbCl₂ = 278.106 g/mol
1 mol of CuCl₂ ------------> 1 mol of PbCl₂
Converting to mass:
134.452 g of CuCl₂ ----------------- 278.106 g of PbCl₂
11.95 g of CuCl₂ ---------------- x g
134.452x = 3323.3667
x = 24.72 g of PbCl₂
The maximum amount of precipitate that could be formed is 24.72 g.
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