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mrs_skeptik [129]

3 years ago

7

How many molecules of o2 are required to form 13.5 g of water vapor?

1 answer:

Viefleur [7K]3 years ago

3 0

13.5g H2O = 13.5/(2+16) = 0.75 mol H2O

Since for every O2 you can form 2 H2O, 0.75/2 = 0.375 mol of O2 will be needed to form 0.75 mol of H2O.

Converting from moles to molecules:

0.375 mol * 6.022 * 10^23 = 2.26 * 10 ^23 molecules

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Answer:

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3 years ago

The following physical constants are for water, H2O.

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Answer:

$Q\approx6.4~kJ$

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

$Q_1=m.C_s.\Delta T$

here;

mass, m = 10 g

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$Q_1=10\times2.09\times 5$

$Q_1=104.5~J$

Amount of heat required to melt the ice at 0°C:

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where, $\Delta H_{fus}=6020~J/mol$

we know that no. of moles is = (wt. in gram) $\div$ (molecular mass)

$Q_2=\frac{10}{18} \times 6020$

$Q_2=3344.44~J$

Now, the heat required to bring the water to 70°C from 0°C:

$Q_3=m.C_L.\Delta T$

specific heat of water, $C_L=4.18~J/g/^oC$

change in temperature, $\Delta T=(70-0)=70^oC$

$Q_3=10\times 4.18\times 70$

$Q_3=2926~J$

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

$Q=Q_1+Q_2+Q_3$

$Q=104.5+3344.44+2926$

$Q=6374.94~J$

$Q\approx6.4~kJ$

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Answer:

the answer is A) carbon

Have a nice day and I hope it helped!

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Explanation:

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4 years ago