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3 years ago

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How many grams of water are made from the reaction of 16.0 grams of oxygen gas?

1 answer:

laiz [17]3 years ago

4 0

Assuming hydrogen gas is in excess, calculate the moles of oxygen

<span>n(O2) = 16.0 g / 32.0 g/mol = 0.500 mol </span>

<span>Calculate the moles of water using the mole ratio; the ratio of H2O to O2 is 2:1 or the moles of H2O is twice the moles of O2 </span>

<span>n(H2O) = 2 x n(O2) = (2)(0.500) = 1.00 mol </span>

<span>Calculate the mass of H2O by multiplying the moles by the molar mass of water </span>

<span>mass = (1.00 mol)(18.0 g/mol) = 18.0 g H2O </span>

<span>Ans: A)

</span>

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Dung dich NaCl 0.9% có 0.9g NaCl trong 100 mL dung dịch

Alenkasestr [34]

Answer:

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3 years ago

Consider the reaction below. Initially the concentration of SO2Cl2 is 0.1000 M. Solve for the equilibrium concentration of SO2Cl

tensa zangetsu [6.8K]

Answer:

$[SO_2Cl_2] = 0.09983 M$

Explanation:

Write the balance chemical equation ,

$SO_2Cl_2((g) = SO_2(g) + Cl_2(g)$

initial concenration of $SO_2Cl_2((g) =0.1M$

lets assume that degree of dissociation= $\alpha$

concenration of each component at equilibrium:

$[SO_2Cl_2] = 0.1-0.1\alpha$

$[SO_2] = 0.1\alpha$

$[Cl_2] = 0.1\alpha$

$Kc =\frac{0.1\alpha \times 0.1\alpha}{0.1-0.1\alpha}$

$Kc =\frac{0.1\alpha \times \alpha}{1-\alpha}$

as $\alpha$ is very small then we can neglect $1-\alpha$

therefore ,

$Kc ={0.1\alpha \times \alpha}$

$\alpha =\sqrt{\frac{Kc}{0.1}}$

$\alpha = 1.73 \times 10^{-3}$

Eqilibrium concenration of $[SO_2Cl_2] = 0.1-0.1\alpha = 0.1-0.1\times 0.00173$

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3 years ago

When driving your car, you hear a squeal every time you press the brakes. You decide the problem is that there's a mouse trapped

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3 years ago

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Answer: Percent recovery is 47.34 %

Explanation:

Percent yield is defined as the ratio of experimental yiled to theoretical yield in terms of percentage.

${\text{ percent yield}}=\frac{\text{amount recovered}}{\text{total amount}}\times 100$

Putting in the values we get:

${\text{ percent yield}}=\frac{0.732}{1.546}\times 100=47.34\%$

Therefore, the percent recovery is 47.34 %

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