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1 year ago

8. 00 g of ethane gas, c2h6, is burned in oxygen. What volume of carbon dioxide gas is produced at 1. 00 atm and 25. 0°c?.

Chemistry

1 answer:

Sonbull [250]1 year ago

8 0

Answer:

Explanation:1 g CH

T = 350 K, P = 1 atm

PV=nRT=

w×R×T

1×V=

×0.0821×350

V=0.495 L

V=495cm

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A site in Pennsylvania receives a total annual deposition of 2.688 g/mof sulfate from fertilizer and acid rain. The ratio by mas

sertanlavr [38]

According to the statement

2.12 x 10^4 lbs pounds of CaCO₃ are needed to neutralize this acid

<h3>What is neutralization?</h3>

A chemical reaction in which an acid and a base react quantitatively with each other is known as neutralization or neutralization. In a water reaction, neutralization ensures that there is no excess of hydrogen or hydroxide ions in the solution.

<h3>According to the given information:</h3>

The equation of the neutralization reaction between H2SO4 and CaCO3.

CaCO3 + H2SO4 → CaSO4 + H2CO3

H2CO3 dissociate to water and carbon dioxide.

CaCO3 + H2SO4 → CaSO4 + H2O + CO2

Now solving for the mass of CaCO3 needed to neutralize the acid.

mass of CaCO3 = 9460 Kg H2SO4 × $\frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO} 4}{98.1 \mathrm{gH}_2 \mathrm{SO}_4} \times \frac{1 \mathrm{~mol} \mathrm{CaCO}\left(\mathrm{O}_3\right.}{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}$ $\times \frac{100.1 \mathrm{~g} \mathrm{CaCO}_3}{1 \mathrm{~mol} \mathrm{CaCO}_3} \times \frac{2.205 \mathrm{lb}}{1000 \mathrm{~g}}$

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1 year ago

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Oxana [17]

The correct answer is 0.014467 M.

Molarity is defined as the number of moles present in a liter solution, that is, number of moles / liter solution.

The molar mass of sodium (Na) is 23.0 g/mol

Thus, 1 mole of Na contains 23.0 g

Now, x moles of Na contains 0.50 g

Moles = 0.50 × 1 / 23.0

Moles = 0.50 / 23.0

= 0.0217 moles of Na

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3 years ago

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topjm [15]

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