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3 years ago

What is the theoretical yield of H2O if 130 g of H2O is produced from 18 g of H2 and an excess of O2?

2 answers:

8 0

You are given 130 grams of H20 and 18 grams of H2. You are required to get the theoretical yield of H2O. Before anything else, you should know the chemical equation of H20 and H2. The balanced chemical reaction is 2H2 + O2 -> 2H2O. The molar mass of H2O is 18 grams per mole and the 2 grams per mole for H2.

18g H2 (1 mol H2/2 g H2)(2 mol H2O/ 2 mol H2)(18 g H2O/1 mol H2O) = 162 g H2O
Thus the theoretical yield for 18g H2 is 162g H2O.

dedylja [7]3 years ago

8 0

The answer to this question on ed is D. 160 just took the test

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Describe the kinetic energy/ movement of the particles in plasma.

exis [7]

kinetic energy is a moving object potential energy is stored energy

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4 years ago

Read 2 more answers

Express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms

nirvana33 [79]

Answer:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$

B = b -a/RT

C = b^2

a = 1.263 atm*L^2/mol^2

b = 0.03464 L/mol

Explanation:

In the given question, we need to express the van der Waals equation of state as a virial expansion in powers of 1/Vm and obtain expressions for B and C in terms of the parameters a and b. Therefore:

Using the van deer Waals equation of state:

$P = \frac{RT}{V_{m}-b } - \frac{a}{V_{m} ^{2} }$

With further simplification, we have:

$P = RT[\frac{1}{V_{m}-b } - \frac{a}{RTV_{m} ^{2} }]$

Then, we have:

$P = \frac{RT}{V_{m} } [\frac{1}{1-\frac{b}{V_{m} } } - \frac{a}{RTV_{m} }]$

Therefore,

$PV_{m} = RT[(1-\frac{b}{V_{m} }) ^{-1} - \frac{a}{RTV_{m} }]$

Using the expansion:

$(1-x)^{-1} = 1 + x + x^{2} + ....$

Therefore,

$PV_{m} = RT[1+\frac{b}{V_{m} }+\frac{b^{2} }{V_{m} ^{2} } + ... -\frac{a}{RTV_{m} }]$

Thus:

$PV_{m} = RT[1 + (b-\frac{a}{RT})\frac{1}{V_{m} } + \frac{b^{2} }{V^{2} _{m} } + ...]$ equation (1)

Using the virial equation of state:

$P = RT[\frac{1}{V_{m} }+ \frac{B}{V_{m} ^{2}}+\frac{C}{V_{m} ^{3} }+ ...]$

Thus:

$PV_{m} = RT[1+ \frac{B}{V_{m} }+ \frac{C}{V_{m} ^{2} } + ...]$ equation (2)

Comparing equations (1) and (2), we have:

B = b -a/RT

C = b^2

Using the measurements on argon gave B = −21.7 cm3 mol−1 and C = 1200 cm6 mol−2 for the virial coefficients at 273 K.

$b = \sqrt{C} = \sqrt{1200} = 34.64[tex]cm^{3}/mol$ [/tex] = 0.03464 L/mol

a = (b-B)*RT = (34.64+21.7)*(1L/1000cm^3)*(0.0821)*(273) = 1.263 atm*L^2/mol^2

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3 years ago

Certain substances have a degreasing effect. Which of the substances below has a degreasing effect in solution?

Eduardwww [97]

Answer:

Sodium Carbonate

Explanation:

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3 years ago

Oxygen always become negatively charged when it forms covalent bonds true or false?

Lelechka [254]

Answer:

I think false

Explanation:

bcs oxygen can create good feeling in our body

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2 years ago

How many grams of sodium chloride are contained in 250.0 g of a 15% NaCl solution?

kramer