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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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Please answer! need it soon
mario62 [17]
Answer to what maybe I can help you
6 0
3 years ago
An object with a mass of 0.225 kg and density of 2.89 g/cm3 measures 34mm in length and 46mm in width. what is the hight of the
Hitman42 [59]
4.15760869565 cm or 41.5760869565
8 0
3 years ago
⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a
STALIN [3.7K]

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

<u>For krypton</u>:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

P x V= \frac{m}{MM} x R x T

m/V= \frac{P x MM}{R x T}

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

           = (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

           = 2.81 g/L

<u>For the gas:</u>

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = \frac{dx R x T}{P}

         = (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

         = 63.99 g/mol ≅ 64 g/mol

<u>For the O₂</u>:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒  m = \frac{P x V x MM}{R x T}= (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

                         = 15.29 g≅ 16 g

4 0
3 years ago
Read 2 more answers
1.For Structure B the bond between carbon and sulfur is polar or non polar
Inga [223]

Answer:

polar

Explanation:

because carbon and sulfur have different electronegativities, the S=C bond is polar.

The entire molecule is nonpolar however because the dipoles (polar bonds) cancel out due to the geometry of the molecule (linear)

3 0
3 years ago
The titration of Na2CO3 with HCl has the following qualitative profile: a. Identify the major species in solution at points A-F.
exis [7]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

Note: This question is incomplete and lacks very important data to solve this question. But I have found the similar question which shows the profiles about which question discusses. Using the data from that question, I have solved the question.

a) We need to find the major species from A to F.

Major Species at A:

1. Na_{2} CO_{3}

Major Species at B:

1. Na_{2} CO_{3}

2. NaHCO_{3}

Major Species at C:

1. NaHCO_{3}

Major Species at D:

1. NaHCO_{3}

2. H_{2}CO_{3}

Major Species at E:

1. H_{2}CO_{3}

Major Species at F:

1. H_{2}CO_{3}

b) pH calculation:

At Halfway point B:

pH = pKa_{1} + log[CO_{3}.^{-2}]/[HCO_{3}.^{-1}]

pH = pKa_{1} = 6.35

Similarly, at halfway point D.  

At point D,

pH = pKa_{2} + log [HCO_{3}.^{-1}]/[H2CO_{3}]

pH = pKa_{2} = 10.33

8 0
3 years ago
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