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3 years ago

A bacteria cell could best be classified as a(n) _____ cell. A. prokaryotic B. eukaryotic C. animal D. plant

Chemistry

2 answers:

Yanka [14]3 years ago

7 0

A. prokaryotic
hope it helped

kicyunya [14]3 years ago

7 0

Hello,

A. prokaryotic

Hope this helps

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Difference between boiling point and freezing point

Paladinen [302]

Answer:

Lower vapor pressures of a substance has an obvious effect on a boiling point. The freezing point the rate at which the solid melts is equal to the rate which freezes.

Explanation:

In practice, small differences between these quantities can be observed. It is difficult, if not impossible, to heat a solid above its melting point because the heat

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4 years ago

How many pints are in a 48 cups?

Feliz [49]

1 cup = 2 pints

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3 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

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None of above. only temperature stays constant.

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3 years ago

Has a definite volume and shape

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3 years ago

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