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3 years ago

I need help!!!!!!!!!!!!!!

Chemistry

1 answer:

kkurt [141]3 years ago

3 0

Answer:

Here is the answer

Explanation:

have higher melting and boiling temperatures than the reactants

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Which of the following is a salt? A. O2 B. C2H2 C. C6H12O6 D. NaCl

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O2 would be your answer.

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4 years ago

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.The vapor pressure of a volatile liquid can be determined by slowly bubbling a known volume of gas through it at a known temper

Elis [28]

Explanation:

It is given that the initial mass of benznene is 7.9286 g

Mass of benzene left = 5.9987 g

So, mass of benzene with which gas get saturated will be calculated as follows.

= 7.9286 g - 5.9987 g = 1.9299 g

Therefore, moles of benzene with which gas get saturated = $\frac{mass}{ molar mass}$

= $\frac{1.9299 g}{78.112 g/mol}$

= 0.0247 moles

Temperature = $27.3^{o}C$ = 27.3 + 273.15 = 300.45 K

Volume = 5.01 L

So, according to ideal gas equation PV = nRT

Putting the given values into the ideal gas equation as follows.

PV = nRT

$P \times 5.01 L$ = $0.0247 mol \times 62.36 torr-liter/mol K \times 300.45 K$

P = $\frac{462.781 torr-liter}{5.01 L}$

= 92.371 torr

Hence, we can conclude that vapor pressure of benzene is 92.371 torr.

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3 years ago

We're doing a lab in chemistry and we have to inflate a ziplock bag. So we need to find the volume of a ziplock bag and i have n

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4 years ago

Two hundred kg of liquid contains 30% butane, 40% pentane, and the rest hexane (mass %) Determine: The mole fraction composition

OlgaM077 [116]

Answer:

The mole fraction composition of the liquid is :

Mole fraction of butane, pentane and hexane are 0.3638,0.3908 and 0.2454 respectively.

Explanation:

Mass of the liquid mixture = 200 g

Percentage of butane = 30%

Mass of butane = $\frac{30}{100}\times 200 g=60 g$

Moles of butane = $n_1=\frac{60 g}{58 g/mol}=1.0345 mol$

Percentage of pentane= 40%

Mass of pentane= $\frac{40}{100}\times 200 g=80 g$

Moles of pentane= $n_2=\frac{80 g}{58 g/mol}=1.1111 mol$

Percentage of hexane = 100% - 30% - 40% = 30%

Mass of hexane = $\frac{30}{100}\times 200 g=60 g$

Moles of hexane = $n_2=\frac{60 g}{86 g/mol}=0.6977 mol$

Mole fraction of butane, pentane and hexane : $\chi_1, \chi_2 \& \chi_3$

$\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{1.0345 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3638$

$\chi_2=\frac{n_2}{n_1+n_2+n_3}=\frac{1.1111 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.3908$

$\chi_3=\frac{n_1}{n_1+n_2+n_3}=\frac{0.6977 mol}{1.0345 mol+1.1111 mol+0.6977 mol}=0.2454$

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4 years ago

For an electron in an atom to change from the ground state to an excited state,

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Answer: some sort of reaction

Explanation: science

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