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Dennis_Churaev [7]
3 years ago
15

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:

No

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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To hit someone on the head

Explanation:

self protection

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C.) 3 is the correct answer

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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} <em>(1)</em>

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}<em>(2)</em>

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

<em>Vapour pressure of cyclohexane at 50°C is 490torr</em>

And for benzene:

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

P^0_{B}=90torr

<em>Vapour pressure of benzene at 50°C is 90torr</em>

3 0
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.
4 0
3 years ago
2. The density of helium is 1.78 X 104 g/cm. What is this<br> density in Dg/um??
Zepler [3.9K]

Answer:

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

Explanation:

Given,

The density of Helium is 1.78\times 10^4\ g/cm^3

We need to find the density in Dg/μm

We know that,

1 g = 10 dg

1 cm³ = 10¹² μm³

So,

d=1.78 \times 10^4\ g/cm^3\\\\=1.78 \times 10^4\times \dfrac{10\ dg}{10^{12}\ \mu m^3}\\\\=1.78\times 10^{-7}\ Dg/\mu m^3

So, the density of Helium is equal to 1.78\times 10^{-7}\ Dg/\mu m^3.

4 0
2 years ago
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