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3 years ago

What is the mole fraction of o2 in a mixture of 15.1 g of o2, 8.19 g of n2, and 2.47 g of h2?

1 answer:

3 0

The solution is
(15.1 g O2) / (31.99886 g O2/mol) = 0.47189 mol O2 (8.19 g N2) / (28.01344 g N2/mol) = 0.29236 mol N2 (2.48 g H2) / (2.01588 g H2/mol) = 1.2302 mol H2
(0.47189 mol O2) + (0.29236 mol N2) + (1.2302 mol H2) = 1.99445 mol total
(0.47189 mol O2) / (1.99445 mol total) = 0.237 for O2

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What else is produced during the combustion of butane, C4H10?<br><br> 2C4H10 + 13O2 → <br> + 10H2O

Neko [114]

<h3>Answer:</h3>

8CO₂

<h3>Explanation:</h3>

We are given;

Butane, C₄H₁₀

Butane is a hydrocarbon in the homologous series known as alkane.

We are required to determine the other product produced in the combustion of butane apart from water.

We know that the complete combustion of alkane yields carbon dioxide and water.
Therefore, combustion of butane will yield carbon dioxide and water.

The balanced equation for the complete combustion of butane will be;

2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O

8 0

3 years ago

Read 2 more answers

1. Name the compound present in a sample of dry air.

aev [14]

Answer:

not air

Explanation:

I think so

hahaha

7 0

2 years ago

Round to 4 significant figures.<br> 0.007062

lapo4ka [179]

Answer:

Hey there!

This is already rounded to four significant figures!

Zeroes after the decimal but before the 7 don't count, and 7, 0, 6, and 2 count as significant figures.

So, the answer would be 0.007062.

Let me know if this helps :)

8 0

2 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

The enzyme urease catalyzes the breakdown of urea in the body. Urease breaks urea down to 2NH3+CO2. This is an example of a hydr

Alika [10]

Answer:

Look at the picture.

Explanation:

On stage one binding of a substrate occurs (and also the geometry of active site may change) and water comes to the site. On stage two the hydrolisis takes place and on stage 3 products deabsorb from the enzyme.

8 0

2 years ago