Gold forms a substitutional solid solution with silver. Compute the weight percent of gold that must be added to silver to yield analloy that contains 7.5 x 1021 Au atoms per cubic centimeter. The densities of pure Au and Ag are 19.32 and 10.49 g/cm³,respectively. The atomic weights for gold and silver are 196.97 and 107.87 g/mol, respectively.i 16.39wt%

The hydronium ion concentration of a 0.20 M solution of hypochlorous acid with a Ka of 3.5 x 10⁻⁸ is 8.4 x 10⁻⁵ M, and the pH is approximately 4.08.

To find the hydronium ion concentration and pH of a 0.20 M solution of hypochlorous acid (HOCl) with a Ka of 3.5 x 10⁻⁸, we need to use the equation for the acid dissociation constant (Ka):

Ka = [H3O+][OCl-]/[HOCl]

We can assume that the dissociation of HOCl is small enough to neglect the concentration of [OCl-], so we can simplify the equation to:

Ka = [H₃O+][OCl-]/[HOCl] ≈ [H₃O+]²/[HOCl]

Rearranging the equation, we get:

[H₃O+]² = Ka x [HOCl] = 3.5 x 10⁻⁸ x 0.20 = 7.0 x 10⁻⁹

Taking the square root of both sides, we get:

[H₃O+] = √(7.0 x 10⁻⁹) = 8.4 x 10⁻⁵ M

To find the pH, we can use the definition of pH:

pH = -log[H₃O+]

Plugging in the value we found for [H₃O+], we get:

pH = -log(8.4 x 10⁻⁵) ≈ 4.08

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The percent yield gives you an idea of how efficient the reaction was. If the percent yield is high, then the reaction was efficient and you obtained a lot of product. If the percent yield is low, then the reaction was not efficient and you did not obtain as much product as you should have.

To find the experimental yield, you need to calculate the actual amount of calcium oxalate that you obtained from the experiment. In this case, you weighed the stone and found it to be 12.993 grams. So, the experimental yield of calcium oxalate is also 12.993 grams.

The theoretical yield is the amount of calcium oxalate that you would expect to obtain if the reaction was 100% efficient. To calculate the theoretical yield, you need to use the balanced chemical equation that you found to get 5 grams of calcium oxalate. If you know the amount of the reactant that you used in the experiment, you can calculate the theoretical yield using stoichiometry.

Finally, to calculate the percent yield, you divide the experimental yield by the theoretical yield and multiply by 100%. So, in this case, you would divide 12.993 by the theoretical yield that you calculated, and then multiply the result by 100% to get the percent yield. The percent yield gives you an idea of how efficient the reaction was. If the percent yield is high, then the reaction was efficient and you obtained a lot of product. If the percent yield is low, then the reaction was not efficient and you did not obtain as much product as you should have.

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The coefficient of the dichromate ion (Cr2O72-) when the following equation is balanced in acidic solution is 1.

Fe2+ + Cr2O72- → Fe3+ + Cr3+

Step-by-step explanation:

Step 1. Balance the chromium atoms by adding a coefficient of 2 in front of Cr3+:
Fe2+ + Cr2O72- → Fe3+ + 2Cr3+

Step 2. Balance the iron atoms by adding a coefficient of 6 in front of Fe2+ and Fe3+:
6Fe2+ + Cr2O72- → 6Fe3+ + 2Cr3+

Step 3. Balance the oxygen atoms by adding a coefficient of 14 in front of the water molecule (H2O) on the right side of the equation:
6Fe2+ + Cr2O72- → 6Fe3+ + 2Cr3+ + 14H2O

Step 4. Balance the hydrogen atoms by adding a coefficient of 22 in front of the hydrogen ion (H+) on the left side of the equation:
6Fe2+ + Cr2O72- + 22H+ → 6Fe3+ + 2Cr3+ + 14H2O

So, the balanced equation is:

6Fe2+ + Cr2O72- + 22H+ → 6Fe3+ + 2Cr3+ + 14H2O

The coefficient of the dichromate ion (Cr2O72-) in the balanced equation is 1.

Therefore, the correct answer is A) 1.

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The pressure of the gas contains 1.75 gram of H₂ and 8.25 grams of Argon is calculated as 7.89 atm.

Pressure of a gas is defined as the force per unit area exerted by gas molecules on the walls of container that encloses them.

As we know ; PV = nRT

For hydrogen gas (H₂):

Molar mass of H₂ = 2.016 g/mol

Number of moles of H₂ = 1.75 g / 2.016 g/mol = 0.8678 mol

For argon gas (Ar):

Molar mass of Ar = 39.948 g/mol

Number of moles of Ar = 8.25 g / 39.948 g/mol = 0.2066 mol

Total number of moles of gas = 0.8678 mol + 0.2066 mol = 1.0744 mol

T = 27.0°C + 273.15 = 300.15 K

P * 3.0 L = 1.0744 * 0.0821 * 300.15

P = (1.0744 * 0.0821 * 300.15 K) / 3.0

P = 7.89 atm

Therefore, the pressure of the gas is 7.89 atm.

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A buffer system can resist changes in pH when small amounts of acid or base are added, by neutralizing them through the reaction of the weak acid and its conjugate base or the weak base and its conjugate acid.

When a strong acid or strong base is added to a buffer system, the pH of the solution is maintained at a relatively constant value. This is because a buffer system contains a weak acid and its conjugate base or a weak base and its conjugate acid, which can neutralize small amounts of added acid or base.

For example, if a strong acid such as hydrochloric acid (HCl) is added to a buffer system containing acetic acid (CH3COOH) and its conjugate base acetate (CH3COO-), the acetate ion will react with the H+ ions from the HCl to form acetic acid, which will prevent the pH of the solution from dropping significantly:

CH3COO- + H+ -> CH3COOH

On the other hand, if a strong base such as sodium hydroxide (NaOH) is added to the same buffer system, the acetic acid will react with the OH- ions from the NaOH to form acetate ion, which will prevent the pH of the solution from rising significantly:

CH3COOH + OH- -> CH3COO- + H2O

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A 50.0-g sample of liquid water at 25.0 °C is mixed with 23.0 g of water at 79.0 °C. The final temperature of the water is 42.0 °C.

At the point when two substances at various temperatures are blended, they will arrive at warm balance, and that implies they will have a similar temperature. To find the last temperature of the water, we can utilize the standard of protection of energy, which expresses that the all out energy in a shut framework stays consistent.

For this situation, we can accept that the two examples of water are the framework, and there is no intensity move to or from the climate. In this way, the intensity lost by the high temp water (q1) will be equivalent to the intensity acquired by the virus water (q2):

q1 = q2

The intensity lost or acquired by a substance can be determined utilizing the recipe:

q = m * c * ΔT

where q is the intensity lost or acquired, m is the mass of the substance, c is its particular intensity limit, and ΔT is the adjustment of temperature.

Utilizing this recipe, we can work out the intensity lost by the high temp water and the intensity acquired by the virus water:

q1 = 23.0 g * 4.184 J/g°C * (79.0°C - temperature)

q2 = 50.0 g * 4.184 J/g°C * (temperature - 25.0°C)

where T is the last temperature of the water.

Comparing q1 and q2, we get:

23.0 g * 4.184 J/g°C * (79.0°C - temperature) = 50.0 g * 4.184 J/g°C * (temperature - 25.0°C)

Addressing for temperature, we get:

temperature = (23.0 g * 4.184 J/g°C * 79.0°C + 50.0 g * 4.184 J/g°C * 25.0°C)/(23.0 g * 4.184 J/g°C + 50.0 g * 4.184 J/g°C)

temperature = 42.0°C

Accordingly, the last temperature of the water is 42.0 °C (choice E).

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The work done for the expansion of CO₂ from 1.0 to 3.0 liters against a pressure of 1.0 atm at constant temperature is -2.0 L*atm.

To calculate the work for the expansion of CO₂ from 1.0 to 3.0 liters against a pressure of 1.0 atm at constant temperature, you can use the formula for work done during an isothermal (constant temperature) expansion:

W = -P * ΔV

where:
- W is the work done
- P is the constant external pressure (1.0 atm in this case)
- ΔV is the change in volume (final volume - initial volume)

Step 1: Calculate the change in volume (ΔV)
ΔV = V_final - V_initial
ΔV = 3.0 L - 1.0 L
ΔV = 2.0 L

Step 2: Substitute the values into the formula
W = -1.0 atm * 2.0 L

Step 3: Calculate the work done
W = -2.0 L*atm

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The solubility product expression for CaF2 is:

Ksp = [Ca2+][F-]2

We know that for every mole of CaF2 that dissolves, one mole of Ca2+ and two moles of F- ions are produced. Let x be the molar solubility of CaF2 in water.

CaF2(s) ⇌ Ca2+(aq) + 2F-(aq)

At equilibrium, the molar solubility of CaF2 can be expressed as:

Ksp = [Ca2+][F-]2

4.0 x 10-8 = (x)(2x)2

4.0 x 10-8 = 4x3

x3 = 1.0 x 10-8

x = 1.0 x 10-8 mol/L

The solubility of CaF2 in water is 1.0 x 10-8 mol/L.

To convert this to g/L, we need to multiply by the molar mass of CaF2:

(1.0 x 10-8 mol/L) x (78.07 g/mol) = 7.81 x 10-7 g/L

Therefore, the solubility of CaF2 in water is 7.81 x 10-7 g/L.

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While the binding of O2 to myoglobin as a function of po2 is described by a simple hyperbolic curve, the binding to hemoglobin is described by a more complex sigmoidal curve.

Myoglobin is a monomeric protein that binds to a single molecule of O2 with high affinity. Its hyperbolic binding curve shows that at low po2, the binding of O2 to myoglobin is rapid and efficient, and as po2 increases, the saturation of myoglobin with O2 approaches 100%.

In contrast, hemoglobin is a tetrameric protein that binds to four molecules of O2 cooperatively. Its sigmoidal binding curve reflects this cooperative binding, showing that at low po2, hemoglobin has low affinity for O2, but as po2 increases, the binding affinity increases due to conformational changes in the protein structure.

The sigmoidal curve allows hemoglobin to pick up O2 in the lungs where po2 is high and release it in the tissues where po2 is low, maximizing oxygen delivery to the body.

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Radium is a radioactive element that decays through alpha decay, which involves the emission of alpha particles from nucleus. This decay process leads to the formation of new element, which is lighter than original element.

Half-life is a term used in radioactive decay to describe the amount of time it takes for half of the sample of radioactive material to decay.

The half-life of radium is 1600 years, which means that it takes 1600 years for half of the original amount of radium to decay. This decay process releases alpha particles and other types of radiation, which can pose significant danger to human health.

Alpha particles are particularly dangerous because they can cause significant damage to living tissue if they are ingested or inhaled.

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The balanced chemical reaction between lithium and nitrogen is;

6Li(s) + N₂(g) → 2Li₃N(s)

From the equation above, we can deduce that; 6 moles of lithium reacts with 1 mole of nitrogen gas. The molar volume of a gas at STP is 22.4 L/mol.

Therefore, at STP, 1 mole of a gas occupies a volume of 22.4 L.

So, 55.2 mL of nitrogen gas at STP is;

55.2/1000 = 0.0552 Liters.

Number of moles of nitrogen gas = volume of gas at STP/STP volume per mole of gas:

0.0552/22.4 = 2.46 × 10⁻³ moles.

Lithium required to react completely with the nitrogen gas:

6 × (2.46 × 10⁻³) = 0.0148 g (approx. 0.015 g).

Therefore, the mass of lithium required to react completely with 55.2 mL of N₂ gas at STP is approximately 0.015 g.

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One approach is to use a lower concentration of NaBH₄, which reduces the amount of hydrogen gas produced.

To minimize hydrogen evolution when using NaBH₄ (sodium borohydride) as a reducing agent in water, you can:

1. Lower the reaction temperature: Conducting the reaction at lower temperatures can reduce the rate of hydrogen gas formation.

2. Use a less reactive reducing agent: If possible, substitute NaBH₄ with a milder reducing agent to limit hydrogen gas production.

3. Control pH: Maintain a slightly acidic pH (around 5) during the reaction to optimize NaBH₄ stability and minimize hydrogen evolution.

4. Add a complexing agent: Utilize complexing agents like ethylenediaminetetraacetic acid (EDTA) to stabilize NaBH₄ and reduce hydrogen gas formation.

5. Use a catalyst: Introduce a catalyst, such as nickel, to facilitate the reduction process and limit the side reaction leading to hydrogen evolution.

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The C-Cl bond in CH3Cl is formed by the overlap of a hybridized orbital on carbon with an unhybridized p orbital on chlorine.

In CH3Cl, carbon is sp3 hybridized, meaning it has four hybridized orbitals that are arranged in a tetrahedral geometry around the carbon atom.

These hybridized orbitals are formed by the combination of one s orbital and three p orbitals from the carbon atom. The hybridization allows the carbon atom to form four single covalent bonds with other atoms.

Chlorine, on the other hand, has an unhybridized p orbital available for bonding. This unhybridized p orbital on the chlorine atom overlaps with one of the sp3 hybridized orbitals on the carbon atom to form the C-Cl bond in CH3Cl.

This type of overlap is called a "sigma bond" and is formed when two orbitals overlap end-to-end along the axis between the two atoms.

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In order of increasing oxidation level b. I < III < II < IV

Explanation:

The oxidation level of a compound is determined by the number of bonds it has to more electronegative atoms (such as oxygen or halogens) and the number of bonds it has to less electronegative atoms (such as carbon or hydrogen).

In CH3COOH (acetic acid), the carbon has one bond to oxygen and two bonds to hydrogen, giving it an oxidation level of +3.

In CH3CH3 (ethane), both carbons have four bonds to hydrogen and no bonds to more electronegative atoms, giving it an oxidation level of 0.

In CH3CHO (acetaldehyde), the carbon has one bond to oxygen, one bond to hydrogen, and one bond to another carbon, giving it an oxidation level of +1.

In CH2=CH2 (ethylene), both carbons have two double bonds to each other and no bonds to more electronegative atoms, giving it an oxidation level of 0.

Therefore, the compounds can be ranked in order of increasing oxidation level as: I < III < II < IV, which is answer choice b.

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Charles's law states that at a constant pressure, the volume of a gas is directly proportional to its absolute temperature. In order to verify this law, an experiment can be conducted where the temperature and volume of a gas are measured at different values. If the results show that the volume increases as the temperature increases and vice versa, then Charles's law is verified.

If your experimental results show a consistent increase in volume as the temperature increases, or a decrease in volume as the temperature decreases, then your results verify Charles's Law. However, if there is no clear relationship between volume and temperature, or the relationship is inconsistent, your results may not support Charles's Law. Keep in mind that experimental errors, such as measurement inaccuracies or equipment issues, can impact your results.

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The form of carbamate that is used mainly as an herbicide is carbaryl.

Carbaryl is a carbamate pesticide that is primarily used to control pests and unwanted plants in agricultural settings. It is a chemical compound made from carbamic acid that replaces one or more hydrogen atoms by other functional groups.

The carbamate insecticide carbaryl was initially approved for use on cotton in the United States in 1959. Today, carbaryl is a popular broad-spectrum pesticide used in residential, commercial, and agricultural settings for the development of decorative plants.

Sevin is the most popular trade name for carbaryl, though it goes by several others as well.

The continued registration and safe use of this insecticide have all been supported by a substantial body of toxicology, environmental fate, residue, and monitoring data that have been compiled over time and combined with actual use experience.

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To calculate the equilibrium constant, kp, for the given reaction at 1 atm and 25°C, we need to use the standard Gibbs free energy change (ΔG°) and the gas constant (R). From the table of thermodynamic data, we have:

ΔG° = -142.7 kJ/mol (for the reaction 2O3(g) ⇌ 3O2(g) at 1 atm and 25°C)
R = 8.314 J/mol·K
We can use the following equation to calculate the equilibrium constant, kp:

ΔG° = -RTln(kp)

where T is the temperature in Kelvin (25°C = 298 K).

Substituting the values, we get:

-142.7 kJ/mol = -(8.314 J/mol·K × 298 K) ln(kp)

Simplifying the equation, we get:

ln(kp) = 54.16

Taking the exponential of both sides, we get:

kp = e^(54.16)

kp = 1.25 × 10^23 (approx)

Therefore, the equilibrium constant, kp, for the reaction 2O3(g) ⇌ 3O2(g) at 1 atm and 25°C is approximately 1.25 × 10^23.
To calculate the equilibrium constant (Kp) for the given reaction (2O3(g) ⇌ 3O2(g)) using thermodynamic data at 1 atm and 25°C, you'll need the standard Gibbs free energy change (ΔG°) values for the reactants and products.

1. Look up the ΔG°f values for O3(g) and O2(g) in a table of thermodynamic data at 1 atm and 25°C.

2. Use the stoichiometric coefficients to calculate the ΔG° for the reaction:

ΔG°(reaction) = Σ[ΔG°f(products)] - Σ[ΔG°f(reactants)]

ΔG°(reaction) = [3 × ΔG°f(O2)] - [2 × ΔG°f(O3)]

3. Use the ΔG°(reaction) value to calculate the equilibrium constant, Kp, using the following equation:

ΔG°(reaction) = -RT ln(Kp)

Where R is the gas constant (8.314 J/mol K), T is the temperature in Kelvin (298K), and ln(Kp) is the natural logarithm of the equilibrium constant.

4. Solve for Kp:

Kp = exp(-ΔG°(reaction) / RT)

Plug in the values and calculate Kp for the given reaction.

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the real concentration of the 0.3 M hydrogen peroxide solution is 6.0 M. To find the real concentration of the 0.3 M hydrogen peroxide solution, we need to use the equation: M1V1 = M2V2

where M1 is the initial concentration of the solution, V1 is the initial volume of the solution, M2 is the real concentration of the solution, and V2 is the volume of the solution used.
Substituting the given values, we get:
0.3 M x V1 = M2 x 5.0 mL
We don't know the initial volume of the solution (V1), so we need to rearrange the equation to solve for M2:
M2 = (0.3 M x V1) / 5.0 mL
Let's assume that the initial volume of the solution was 100 mL. Substituting this value, we get:
M2 = (0.3 M x 100 mL) / 5.0 mL
M2 = 6.0 M
Therefore, the real concentration of the 0.3 M hydrogen peroxide solution is 6.0 M.

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The net ionic equation for the reaction between aqueous solutions of sn(c2h3o2)2 and koh can be written as follows:
Sn(C2H3O2)2(aq) + 2OH-(aq) → Sn(OH)2(s) + 2C2H3O2-(aq)

The equation represents a chemical reaction between two ionic compounds: Sn(C2H3O2)2 and KOH. When these two compounds are mixed, they undergo a double replacement reaction, where the cations (positive ions) of the two compounds swap places, resulting in the formation of two new compounds: tin (II) hydroxide (Sn(OH)2) and potassium acetate (C2H3O2-).

The net ionic equation for this reaction only shows the species that participate in the reaction, which are the ions that are present before and after the reaction takes place. This equation is written by canceling out the spectator ions, which are the ions that do not participate in the reaction and are present on both sides of the equation.

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To solve this problem, we need to use the equations for the dissociation of a diprotic acid.

The first dissociation is:
H2L ⇌ H+ + HL-

The equilibrium constant for this reaction is Ka1 = [H+][HL-]/[H2L].

We are given that Ka1 = 1.0x10-3.

The second dissociation is:
HL- ⇌ H+ + L2-

The equilibrium constant for this reaction is Ka2 = [H+][L2-]/[HL-].

We are given that Ka2 = 1.0x10-11.

We are also given that the solution contains H2L, HL, L-, and H. This means that some of the H2L will dissociate to form HL- and H+, and some of the HL- will dissociate to form L2- and H+. We can set up an ICE table for each dissociation:

For the first dissociation:
H2L ⇌ H+ + HL-
I   0.65        0         0
C  -x           +x        +x
E  0.65-x     x         x

For the second dissociation:
HL- ⇌ H+ + L2-
I    x             0         0
C   -y            +y        +y
E    x-y        y         y

We can use the equilibrium constants to set up expressions for x and y:
Ka1 = [H+][HL-]/[H2L] = x^2/(0.65-x)
Ka2 = [H+][L2-]/[HL-] = y^2/x

We can solve for x and y using these expressions and the quadratic formula. However, since Ka2 is much smaller than Ka1, we can assume that most of the HL- will not dissociate to form L2- and H+. This means that y will be much smaller than x, and we can simplify the expressions to:

Ka1 = [H+][HL-]/[H2L] = x^2/0.65
Ka2 = [H+][L2-]/[HL-] ≈ y^2/x ≈ 0

Solving for x, we get:
x = sqrt(Ka1*[H2L]/0.65) = 1.14x10^-3 M

Now we can use the expression for the pH:
pH = -log[H+] = -log(x) = 2.94

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Saponification is an example of b) ester hydrolysis.

Saponification is a chemical reaction that is commonly used in the production of soap, type of  b) ester hydrolysis. The process involves the reaction of a fat or oil (triglyceride) with a strong alkali, such as sodium hydroxide or potassium hydroxide. The reaction results in the formation of soap, which is a salt of a fatty acid, and glycerol, which is a three-carbon alcohol.

Triglycerides are esters formed from the reaction of glycerol and fatty acids. They are commonly found in animal and vegetable fats and oils. In saponification, the ester bonds present in the triglyceride molecule are hydrolyzed by the strong alkali, resulting in the formation of soap and glycerol.

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Approximately 6.02 x 1025 mole of sand particles would be present in the truck.

A substance's mole is equivalent to 6.022 x 1023 of that substance. (such as atoms, molecules, or ions). The number 6.022 1023 is referred to as "Avogadro's number" or "Avogadro's constant." To translate between particle mass and number, use the mole idea.

Yes, if we know how many moles of sand are in the truck, we can calculate how many sand particles are there.

Avogadro's number, which is the quantity of particles (atoms, molecules, or ions) in one mole of a substance, can be used to calculate the number of particles in a substance. The equivalent of 6.02 x 1023 particles/mol is Avogadro's number.

Number of particles = Avogadro's number x the number of moles

The quantity of sand particles can be determined as follows if the truck contains 100 moles of sand:

100 moles times 6.02 x 1023 particles/mol equals the number of particles.

6.02 x 1025 particles make up the total.

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When bromothymol blue is green, it indicates that the solution is at its transition pH range, which is approximately pH 6.0 to 7.6. At this point, the relative proportions of the acidic form (HIn) and the basic form (In-) are equal. So, for bromothymol blue in its green intermediate color, the ratio of HIn to In- is 1:1.

The relative proportions of HIn and In when bromothymol blue is green depend on the pH of the solution. At a pH of around 6, the equilibrium between the two forms of the indicator is roughly equal, meaning that both HIn and In are present in roughly equal amounts. As the pH shifts towards more acidic conditions, the proportion of HIn increases relative to In, and the color of the indicator shifts towards yellow. Conversely, as the pH shifts towards more basic conditions, the proportion of In increases relative to HIn, and the color of the indicator shifts towards blue. So, to answer your question specifically, we would need to know the exact pH of the solution in order to determine the relative proportions of HIn and In when bromothymol blue is green.

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To determine which substance has the greater value of S, we need to look at the standard molar entropy (S°) values for each substance. a)At 0 K, both N2O and He have a S° of 0 J/mol·K. Therefore, they have the same value of S.

b)At 25°C (298 K), the S° for N2O(g) is 219.7 J/mol·K and for He(g) it is 126.2 J/mol·K. Therefore, N2O(g) has the greater value of S, c. The entropy change (ΔS) for the phase transition from NH3(s) to NH3(l) at 196 K is positive, meaning that the entropy of the system increases as the solid turns into a liquid. Therefore, NH3(l) has a greater value of S than NH3(s).

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During the communist rule in Central Asia, particularly in China, the Soviet Union, and Mongolia, religion experienced significant suppression and control. This period saw a decline in the influence and practice of religion in Central Asia, although some underground religious activities persisted.

During communist rule in Central Asia by China, the Soviet Union, and Mongolia, religion faced significant suppression and restrictions. Communist regimes viewed religion as a threat to their ideology and worked to eradicate it from society. Mosques, temples, and churches were closed, and religious leaders were persecuted, imprisoned, or executed. In China, the Cultural Revolution led to the destruction of many religious artifacts and institutions. The Soviet Union implemented anti-religious campaigns and established state-sponsored atheism. In Mongolia, Buddhism, which was the dominant religion, was banned, and many monasteries were destroyed. Overall, the communist regimes in Central Asia suppressed religion, resulting in a decline in religious practice and influence in the region.

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The given question is related to a specific experiment and asks why a large excess of SCN- was used in the first part of the experiment. Without additional information about the experiment, it is difficult to provide a specific answer.

However, in general, using an excess of one reagent in a chemical reaction can help to ensure that the reaction goes to completion and that all of the limiting reagent is consumed.In the context of a specific experiment, using a large excess of SCN- may have been necessary to drive the reaction towards completion and ensure that all of the Fe3+ ions in the solution were complexed with SCN- to form the red Fe(SCN)2+ complex. Alternatively, it may have been used to minimize interference from other ions or organic compounds in the solution that could have interfered with the reaction or affected the accuracy of the results.Overall, the reason for using a large excess of SCN- in a specific experiment will depend on the experimental conditions and the desired outcome of the experiment. It requires an understanding of the principles of chemical reactions and the properties of the reagents and reactants involved.

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The large peak stretching from 2500-3500 could be due to the O-H stretch from an alcohol or carboxylic acid, and the sharp peak around 3000 could be due to a C-H stretch from sp3 hybridized carbon.

The molecular formula of compound C is C4H8O2. To calculate the Unsaturation Number (UN) or Double-Bond Equivalent (DBE), we use the formula:

DBE = C - (H/2) + (N/2) + 1

In this case, we have 4 carbon atoms, 8 hydrogen atoms, and 0 nitrogen atoms, so the calculation is:

DBE = 4 - (8/2) + (0/2) + 1 = 4 - 4 + 0 + 1 = 1

The Unsaturation Number (UN) or Double-Bond Equivalent (DBE) for compound C is 1.

Based on the UN/DBE you calculated, there could be a double bond in compound C.

Based on the UN/DBE you calculated, there could also be a ring in compound C.

As for the IR spectrum, the large peak stretching from 2500-3500 could be due to the O-H stretch from an alcohol or carboxylic acid, and the sharp peak around 3000 could be due to a C-H stretch from sp3 hybridized carbon.

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the total mass of products when 2.20 g of propane is burned in excess oxygen is  10.20 g

To solve this problem, we need to use stoichiometry. First, we need to balance the chemical equation:

C3H8 + 5O2 -> 3CO2 + 4H2O

From the balanced equation, we can see that for every 1 mole of propane (C3H8) burned, we will produce 3 moles of carbon dioxide (CO2) and 4 moles of water (H2O).

To find the total mass of products, we need to convert the given mass of propane (2.20 g) to moles:

2.20 g C3H8 x (1 mol C3H8/44.1 g C3H8) = 0.050 mol C3H8

Next, we can use the mole ratio from the balanced equation to find the number of moles of CO2 and H2O produced:

0.050 mol C3H8 x (3 mol CO2/1 mol C3H8) = 0.150 mol CO2
0.050 mol C3H8 x (4 mol H2O/1 mol C3H8) = 0.200 mol H2O

Finally, we can convert the moles of CO2 and H2O to grams:

0.150 mol CO2 x (44.01 g CO2/1 mol CO2) = 6.60 g CO2
0.200 mol H2O x (18.02 g H2O/1 mol H2O) = 3.60 g H2O

Therefore, the total mass of products when 2.20 g of propane is burned in excess oxygen is:

6.60 g CO2 + 3.60 g H2O = 10.20 g

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by adding 0.841 L of a 0.266 M solution of KI to a solution of excess Pb(NO₃)₂, we can form 51.58 g of PbI₂.

we need to use stoichiometry and the balanced equation for the reaction between KI and Pb(NO₃)₂:

2KI + Pb(NO₃)₂ → 2KNO₃ + PbI₂

From this equation, we can see that 2 moles of KI react with 1 mole of Pb(NO₃)₂ to form 1 mole of PbI₂. Therefore, we need to first calculate the number of moles of KI present in the solution:

moles of KI = molarity x volume in liters

= 0.266 M x 0.841 L

= 0.223706 moles KI

Since there is excess Pb(NO₃)₂ present, all of the KI will react to form PbI₂. Therefore, we can use the moles of KI to calculate the moles of PbI₂ that will be formed:

moles of PbI2 = 0.223706 moles KI ÷ 2

= 0.111853 moles PbI2

Finally, we can use the molar mass of PbI₂ (461.01 g/mol) to calculate the mass of PbI₂ that will be formed:

mass of PbI₂ = moles of PbI₂ x molar mass

= 0.111853 moles x 461.01 g/mol

= 51.58 g

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