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solniwko [45]
2 years ago
15

Question 3 0 / 1 point How many grams of carbon atoms are present in a sample of C3H8 if there are 4.10 moles of hydrogen atoms

in the sample
Chemistry
1 answer:
hodyreva [135]2 years ago
5 0

Answer:

18.45 g of C

Explanation:

This is a problem of rules of three:

1 mol of C₃H₈ contains 3 moles of C and 8 moles of H

If 8 moles of H are contained in 1 mol of propane

4.10 moles of H are contained in (4.1 . 1) /8 = 0.5125 moles

Now, If 1 mol of propane contains 3 moles of C

0.5125 moles of propane may contain (0.5125 . 3) / 1 = 1.5375 moles of C

Let's convert the moles to mass:

1.5375 mol . 12 g /mol = 18.45 g

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Consider the following reaction: CH3OH(g)⇌CO(g)+2H2(g) Part A Calculate ΔG for this reaction at 25 ∘C under the following condit
kati45 [8]

<u>Answer:</u> The \Delta G of the reaction at given temperature is -12.964 kJ/mol.

<u>Explanation:</u>

For the given chemical reaction:

CH_3OH(g)\rightleftharpoons CO(g)+2H_2(g)

The expression of K_p for the given reaction:

K_p=\frac{(p_{CO})\times (p_{H_2}^2)}{p_{CH_3OH}}

We are given:

p_{CO}=0.140atm\\p_{H_2}=0.180atm\\p_{CH_3OH}=0.850atm

Putting values in above equation, we get:

K_p=\frac{(0.140)\times (0.180)^2}{0.850}\\\\K_p=5.34\times 10^{-3}

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 = 0 J (at equilibrium)

R = Gas constant = 8.314J/K mol

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

K_p = equilibrium constant in terms of partial pressure = 5.34\times 10^{-3}

Putting values in above equation, we get:

\Delta G=0+(8.314J/K.mol\times 298K\times \ln(5.34\times 10^{-3}))\\\\\Delta G=-12963.96J/mol=-12.964kJ/mol

Hence, the \Delta G of the reaction at given temperature is -12.964 kJ/mol.

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When cellular respiration occurs, food energy is converted into chemical energy in the way of ATP, ______ and _____ are the prod
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Carbon dioxide and Water are the products of cellular respiration.
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Based on the equation 3 Cu (s) + 8 HNO3 (aq) 3 Cu(NO3)2 (aq) + 2NO (g) + 4H2O (g) how many grams of Cu would be needed to react
eimsori [14]

Answer:

Mass = 381.28 g

Explanation:

Given data:

Number of moles of HNO₃ = 16 mol

Mass of Cu needed to react with 16 mol of HNO₃ = ?

Solution:

Chemical equation:

3Cu + 8HNO₃    →     3Cu(NO₃)₂ + 4H₂O + 2NO

Now we will compare the moles of Cu with HNO₃ from balance chemical equation.

                     HNO₃         :          Cu

                        8              :         3

                       16             :       3/8×16 = 6

Mass of Cu needed:

Mass = number of moles × molar mass

Mass = 6 mol × 63.546 g/mol

Mass = 381.28 g

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