Solid carbon tetrachloride, CCl4(s), is represented by the diagram above. The attractions between the CCl4 molecules that hold t
1 answer:
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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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:
m/V=
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 =
= (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 = = (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
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.
Major Species at B:
1.
2.
Major Species at C:
1.
Major Species at D:
1.
2.
Major Species at E:
1.
Major Species at F:
1.
b) pH calculation:
At Halfway point B:
pH = pK + log[
]/[H
]
pH = pK = 6.35
Similarly, at halfway point D.
At point D,
pH = pK + log [H
]/[H2
]
pH = pK = 10.33