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3 years ago

How many molecules are in 2.0 moles of carbon dioxide?

Chemistry

1 answer:

zubka84 [21]3 years ago

7 0

Answer:

There are 1.51 x 1024 molecules of carbon dioxide in 2.50 moles of carbon dioxide.

Explanation:

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8_murik_8 [283]

I think it is "Known".

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3 years ago

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50 grams of acetic acid C2H4O2 are dissolved in 200 g of water. Calculate the weight % and mole fraction of the acetic acid in t

Len [333]

$50g$ $of$ $C_2H_4O_2$ $in$ $200g$ $H_2O$

so:

$\frac{50}{200+50}+100\%= \frac{50*100}{250}\%=20\%$

$MW of acid = 2*C+4*H+2*O = 2*12+4*1+2*16=$

$=24+4+32=60g/mol$

so:

$\frac{50g}{60g/mol}\approx0.83moles$

it means that

in $50g$ of acid there is $\approx0.83$ moles of acid

$MW of H_2O = 2*H+O=2*1+16=2+16=18g/mol$

so:

$\frac{200g}{18g/mol}\approx11.11moles$

it means that:

in $200g$ of water there is $\approx11.11$ moles of water

therefore:

$\frac{0.83mol}{11.11mol+0.83mol}= \frac{0.83mol}{11.94mol}=0.069$

So your answers are:

$20\%$

and the mole fraction is:

$0.069$

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3 years ago

5.36 liters of nitrogen gas are at STP. What would be the new volume if we increased the moles from 3.5 moles to 6.0 moles?

Aleksandr [31]

Answer:

$V_2=9.20L$

Explanation:

Hello there!

In this case, according to the given STP (standard pressure and temperature), it is possible for us to realize that the equation to use here is the Avogadro's law as a directly proportional relationship between moles and volume:

$\frac{V_2}{n_2}= \frac{V_1}{n_1}$

In such a way, given the initial volume and both initial and final moles, we can easily compute the final volume as shown below:

$V_2= \frac{V_1n_2}{n_1} \\\\V_2=\frac{5.36L*6.0mol}{3.5mol}\\\\V_2=9.20L$

Best regards!

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3 years ago

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lions [1.4K]

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3 years ago

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kati45 [8]

Answer:

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Explanation:

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