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4 years ago

An equilibrium mixture of O2, SO2 and SO3 contains equal concentrations of SO2 and SO3.

Chemistry

1 answer:

AleksAgata [21]4 years ago

7 0

Answer:

[O₂(g)] = 0.0037M

Explanation:

2SO₂(g) + O₂(g) => 2SO₃(g)

Conc: [SO₂(g)] [O₂(g)] [SO₃(g)] and [SO₂(g)] = [SO₃(g)]

Kc = [SO₃(g)]²/[O₂(g)][SO₂(g)]² => Kc = 1/[O₂(g)] = 270 if [SO₂(g)] = [SO₃(g)]

∴ [O₂(g)] = (1/270)M = 0.0037M

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Ill give the brainliest answer to whoever helps me with this equation

vampirchik [111]

Answer: The percent yield for the $NaBr$ is, 86.7 %

Explanation : Given,

Moles of $FeBr_3$ = 2.36 mol

Moles of $NaBr$ = 6.14 mol

First we have to calculate the moles of $NaBr$

The balanced chemical equation is:

$2FeBr_3+3Na_2S\rightarrow Fe_2S_3+6NaBr$

From the reaction, we conclude that

As, 2 moles of $FeBr_3$ react to give 6 moles of $NaBr$

So, 2.36 moles of $FeBr_3$ react to give $\frac{6}{2}\times 2.36=7.08$ mole of $NaBr$

Now we have to calculate the percent yield for the $NaBr$ .

$\text{Percent yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100$

Experimental yield = 6.14 moles

Theoretical yield = 7.08 moles

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

$\text{Percent yield}=\frac{6.14mol}{7.08mol}\times 100=86.7\%$

Therefore, the percent yield for the $NaBr$ is, 86.7 %

6 0

3 years ago

Each step in the following process has a yield of 90.0 %. CH 4 + 4 Cl 2 ⟶ CCl 4 + 4 HCl CCl 4 + 2 HF ⟶ CCl 2 F 2 + 2 HCl The CCl

Rainbow [258]

Answer:

4.86 moles of HCl

Explanation:

1. First write the balanced chemical equations involved in the process:

$CH_{4}+_4Cl_{2}=CCl_{4}+_4HCl$

$CCl_{4}+_2HF=CCl_{2}+F_{2}+_2HCl$

2. Calculate what amount of $CCl_{4}$ is formed in the first reaction.

$3.00molesCH_{4}*\frac{1molCCl_{4}}{1molCH_{4}}=3.00molesCCl_{4}$

As the yield of each reaction is 90.0%, the amount of $CCl_{4}$ produced is the following:

$3.00molesCCl_{4}*0.90=2.7molesCCl_{4}$

3. Calculate the amount of HCl produced.

$2.7molesCCl_{4}*\frac{2molHCl}{1molCCl_{4}}=5.4molesHCl$

The total amount of HCl produced with a 90.0% yield is:

$5.4molesHCl*0.90=4.86molesHCl$

4 0

3 years ago

A 100.0 mL solution containing 0.864 g of maleic acid (MW=116.072 g/mol) is titrated with 0.276 M KOH. Calculate the pH of the s

Lilit [14]

Answer:

pH = 1.32

Explanation:

H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺

This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:

So first calculate the moles reacted and produced:

n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M

54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH

it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.

moles H₂M left = 0.074 - 0.015 = 0.059

moles HM⁻ produced = 0.015

Using the Henderson - Hasselbach equation to solve for pH:

ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325

Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.

For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.

3 0

3 years ago

What is 8S8 In need for a quick answer

kirza4 [7]

Answer:

8x8 Inc. is a provider of Voice over IP products. 8x8 products include cloud-based voice, contact center, video, mobile and unified communications for businesses.

Explanation:

7 0

3 years ago

Equation 3x-5=-2x+ 10? O 0 x= 5 -5 = x -15 = -5x -5x = 15

Harman [31]

Answer:

are those the answer choice

7 0

4 years ago