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Marina86 [1]
3 years ago
8

C6h12(l)+9O2(g)->6CO2(g)+6h2O(g)

Chemistry
1 answer:
gavmur [86]3 years ago
6 0
The answer is92bc multiply all of it then add h20g
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Answer:

Galileo Galilei

What launched the era of modern science in the 17th century? Modern science began in the 17th century, when the Italian physicist Galileo Galilei revived the Copernican view.

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Explanation:

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Which is a spectator ion involved in the reaction of k2cro4(aq) and ba(no3)2(aq)?
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K₂CrO₄(aq) + Ba(NO₃)₂(aq) = BaCrO₄(s) + 2KNO₃(aq)

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The partial pressure of CH4 is 0.175 atm and that of O2 is 0.250 atm in a mixture of the two gases. What is the mole fraction of
tresset_1 [31]

Answer:

The total number of moles of gas in the mixture is 0.16939.

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Explanation:

Total volume of the mixture = V = 10.5 L

Temperature of the mixture = T = 35°C = 308.15K

Pressure of the mixture = P

Total moles of mixture = n =n_1+n_2

Using an ideal gas equation :

PV=nRT

P\times 10.0 L=n\times 0.0821 atm L/mol K\times 308.15 K

P=2.509 n

Partial pressure of the methane= p_1=0.175 atm

Moles of the methane= n_1

Partial pressure of the oxygen gas= p_2=0.250 atm

Moles of the methane= n_2

Mole fraction  of the methane= \chi_1=\frac{n_1}{n_1+n_2}

Mole fraction  of the oxygen gas= \chi_2=\frac{n_2}{n_1+n_2}

p_1=p\times \chi_1  (Dalton's law)

0.175 atm=P\times \frac{n_1}{n_1+n_2}

0.175 atm=(2.509 n)\times \frac{n_1}{n}

n_1=0.06975 mol

Mass of 0.06975 moles of methane gas :

0.06975 mol × 18 g/mol =1.25550 g

p_2=p\times \chi_2  (Dalton's law)

0.250 atm=P\times \frac{n_2}{n_1+n_2}

0.250 atm=(2.509 n)\times \frac{n_2}{n}

n_2=0.09964 mol

Mass of 0.06975 moles of oxygen gas :

0.09964 mol × 32 g/mol =3.18848 g

Total moles of mixture = n =n_1+n_2

[/tex]=0.06975 mol+0.09964 mol=0.16939 mol[/tex]

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3 years ago
When a solid melts
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When solid melts, the temperature of the substances increases and the substance absorbs heat energy, converting the solid to a liquid.
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A balloon has a volume of 33L at ground level and when it rises to 10,000m the balloon will have a volume of 148 L at 0.5 atm. W
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