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Veronika [31]
2 years ago
5

Solid carbon tetrachloride, CCl4(s), is represented by the diagram above. The attractions between the CCl4 molecules that hold t

he molecules together in the solid state are best identified as
(A) polar covalent bonds
(B) nonpolar covalent bonds
(C) intermolecular attractions resulting from temporary dipoles
(D) intermolecular attractions resulting from permanent dipoles
Chemistry
1 answer:
Ne4ueva [31]2 years ago
7 0

Hi, you've asked an incomplete question. Here's the diagram that completes the question.

Answer:

<u>(B) nonpolar covalent bonds</u>

Explanation:

This structure in the diagram rightly fits the description of a non-covalent bond because there is an equal sharing of electrons of Carbon (C) and Chlorine (Cl).

<em>Remember</em> too that these elements are in their solid-state, hence the CCl4 (carbon tetrachloride) molecules are held strongly together.

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<u>Volume </u><u>of</u><u> 106.9 mL from the concentrated solution should be taken and diluted to 350 </u><u>mL.</u>

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<u>We can use the </u><u>formula.</u><u> </u>

c1v1 =c2v2

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and v2 is the volume of the diluted solution to be prepared

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2 years ago
What properties do protons and electrons share
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Protons and Electrons charges are the exact same size.
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(multiple choice) What is the SI unit of mass?
SpyIntel [72]

Answer:

Gram

Explanation:

The SI unit of mass is grams.

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2 years ago
Sample of gas initially occupies 4.25 L at a pressure of 0.850 atm at 23.0°C. What will the volume be if the temperature is chan
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Answer:  2.34 L

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas = 0.850 atm

P_2 = final pressure of gas = 1.50 atm

V_1 = initial volume of gas = 4.25 L

V_2 = final volume of gas = ?

T_1 = initial temperature of gas = 23.0^oC=273+23.0=293.0K

T_2 = final temperature of gas = 11.5^oC=273+11.5=284.5K

Now put all the given values in the above equation, we get:

\frac{0.850\times 4.25}{293.0K}=\frac{1.50\times V_2}{284.5}

V_2=2.34L

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The volume of a gas at 30◦C and 0.13 atm is 61 mL. What volume will the same gas sample occupy at standard conditions?
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