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3 years ago

Suppose you have a mole of uranium (U) and three moles of hydrogen (H). Do you have

Chemistry

1 answer:

julia-pushkina [17]3 years ago

3 0

Answer:

Explanation:

A mole is a measure of a number of atoms. One mole of Uranium is the same number of atoms as one mole of Hydrogen

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Q4. What are the different ways in which human being use rocks ?

Ghella [55]

Answer:

To hit someone on the head

Explanation:

self protection

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2 years ago

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11111nata11111 [884]

Answer:

C.) 3 is the correct answer

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3 years ago

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Mixtures of benzene and cyclohexane exhibit ideal behavior. A solution was created containing 1.5 moles of liquid benzene and 2.

goblinko [34]

Answer:

Vapour pressure of cyclohexane at 50°C is 490torr

Vapour pressure of benzene at 50°C is 90torr

Explanation:

Using Raoult's law, pressure of a solution is defined by the sum of the product sbetween mole fraction of both solvents and pressure of pure solvents.

$P_{solution} = X_{A}P^0_{A}+X_{B}P^0_{B}$

In the first solution:

$X_{cyclohexane}=\frac{2.5mol}{2.5mol+1.5mol} =0.625$

$X_{benzene}=\frac{1.5mol}{2.5mol+1.5mol} =0.375$

$340torr = 0.625P^0_{A}+0.375P^0_{B}$ (1)

For the second equation:

$X_{cyclohexane}=\frac{3.5mol}{3.5mol+1.5mol} =0.700$

$X_{benzene}=\frac{1.5mol}{3.5mol+1.5mol} =0.300$

$370torr = 0.700P^0_{A}+0.300P^0_{B}$ (2)

Replacing (2) in (1):

$340torr = 0.625P^0_{A}+0.375(1233.3-2.333P^0_{A})$

$340torr = 0.625P^0_{A}+462.5-0.875P^0_{A}$

-122.5torr = -0.250P°A

$P^0_{A} = 490 torr$

Vapour pressure of cyclohexane at 50°C is 490torr

And for benzene:

$370torr = 0.700*490torr+0.300P^0_{B}$

$P^0_{B}=90torr$

Vapour pressure of benzene at 50°C is 90torr

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2 years ago

What happens if a geologist drips a small amount of vinegar (acetic acid) onto a sample of dolomite?

Lapatulllka [165]

Hydrochloric acid on a rock or mineral and watching for bubbles of carbon dioxide gas to be released. The bubbles signal the presence of carbonate minerals such as calcite and dolomite.

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3 years ago

2. The density of helium is 1.78 X 104 g/cm. What is this density in Dg/um??

Zepler [3.9K]

Answer:

$d=1.78\times 10^{-7}\ Dg/\mu m^3$