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Bogdan [553]
3 years ago
10

Consider the reaction 2 CO + O2 → 2 CO2 .What is the percent yield of carbon dioxide

Chemistry
1 answer:
Wewaii [24]3 years ago
7 0

Answer:

Y=50.9\%

Explanation:

Hello,

In this case, given the reaction, we can directly compute the theoretically yielded grams of carbon dioxide, considering the 2:2 molar ratio between carbon monoxide (molar mass = 28 g/mol) and carbon dioxide (molar mass = 44 g/mol) and the initial reacting grams of carbon monoxide in excess oxygen:

m_{CO_2}^{theoretical}=10gCO*\frac{1molCO}{28gCO}*\frac{2molCO_2}{2molCO}*\frac{44gCO_2}{1molCO_2}   =15.7gCO_2

Thus, as only 8 g were actually yielded, we compute the percent yield:

Y=\frac{8g}{15.7g}*100\% \\\\Y=50.9\%

Best regards.

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QUESTION 3 Consider a solution containing 0.80 M NaF and 0.80 M HF. Calculate the moles of HF and the concentration of HF after
Lisa [10]

Answer:

0.056moles HF and 0.70M

Explanation:

When a strong acid is added to a buffer, the acid reacts with the conjugate base.

In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:

NaF + HCl → HF + NaCl

Initial moles of NaF and HF in 60.0mL of solution are:

NaF:

0.0600L × (0.80mol / L)= 0.048 moles NaF

HF:

0.0600L × (0.80mol / L)= 0.048 moles HF

Then, the added moles of HCl are:

0.0200L × (0.40mol / L) = 0.008 moles HCl.

Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:

<em>0.056moles HF</em>

<em></em>

In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:

0.056mol HF / 0.0800L = <em>0.70M</em>

6 0
3 years ago
How many moles are represented by 3.01 - 1023 helium atoms?
Pavlova-9 [17]

Answer:

0.5 mol

Explanation:

Given data:

Number of atoms of He = 3.01 ×10²³

Number of moles = ?

Solution:

Avogadro number:  

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.02 × 10²³ is called Avogadro number.

1 mole = 6.02 × 10²³ atoms

3.01 ×10²³ atoms × 1 mol / 6.02 × 10²³ atoms

0.5 mol

7 0
3 years ago
Question 1 (3 points)
den301095 [7]

Answer:

Explanation:

ΔTemp => 35⁰C(108K) increases to 57.9⁰C(330.9L) => increases volume (Charles Law)

Use the Kelvin Temperature values in a ratio that will increase the original volume.

ΔVol = 6.33L(330.9/108.0) => gives a larger volume. Using 108.0/330.9 would give a smaller volume and would be contrary to what the problem is asking.

ΔPress => 342 mmHg increases to 821 mmHg => decreases volume (Boyles Law)

Use the pressure values in a ratio that will decrease the original volume.

ΔPress = 6.33L(342/821) => gives a smaller volume. Using 821/342 would give a larger volume and would be contrary to what the problem is asking.

Now, putting both ΔTemp together with ΔPress => net change in volume. (Combined Gas Law)

ΔVol = 6.33L(330.9/108.0)(342/821) = 8.08L (final volume of gas).

___________________

This problem can also be worked using the combined gas law equation:

P₁V₁/T₁ = P₂V₂/T₂ => V₂ = P₁V₁T₂/T₁P₂

V₂ = [(342mm)(6.33L)(330.9K)]/[(108K)(821mm)] = 8.08L (final volume of gas)

7 0
3 years ago
Help me on this one too
nlexa [21]

7 is the pancreas and 8 is i can’t see the other choices but for diabetes the glucose levels will be a lot higher

3 0
4 years ago
Read 2 more answers
what is the percent by mass of a solution that contains 30 grams of potassium nitrite in 0.5 kilograms of water?
Makovka662 [10]

Answer:

5.66 %.

Explanation:

<em>mass percent is the ratio of the mass of the solute to the mass of the solution multiplied by 100.</em>

<em />

<em>mass % = (mass of solute/mass of solution) x 100.</em>

<em></em>

mass of potassium nitrite = 30.0 g,

mass of the solution = mass of water + mass of potassium nitrite = 500.0 g + 30.0 g = 530.0 g.

<em>∴ mass % = (mass of solute/mass of solution) x 100</em> = (30.0 g/530.0 g) x 100 = <em>5.66 %.</em>

5 0
3 years ago
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