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3 years ago

True or False the molecule CCl4 contains one double bond and three single bonds.

Chemistry

2 answers:

tensa zangetsu [6.8K]3 years ago

5 0

Answer: False

Explanation: Carbon is the central atom which has valency of 4 and can form 4 bonds to complete its octet. Each chlorine has valency of 1 and can form only one bond to complete its octet. Thus carbon can bond with four chlorine atoms through sharing of electrons to form four single bonds. Thus it cannot contain a double bond.

$C:6:1s^22s^22p^2$

$Cl:17:1s^22s^22p^63s^23p^5$

lisabon 2012 [21]3 years ago

4 0

False: C-Cl is a single bond as Cl is -1 in its valency and C is 4.

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The ability to be maintained at a certain rate or level

Explanation:

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3 years ago

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Mutualisms/Symbiotic relationships

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2 years ago

You conduct an experiment in which you measure the temperature (T) and volume (V) of a mysterious sphere of gas at several diffe

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Volume of the gas is 525 L.

<u>Explanation:</u>

It is given that the volume of the gas divided by the temperature is 1.75.

V/T = 1.75

As per the Charles law, volume is proportional to the temperature.

V ∝ T

V/T = constant

Now we have to find V, and T is given as 300 K.

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V/300 = 1.75

Rearranging the equation to get V as,

V = 1.75×300

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3 years ago

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miss Akunina [59]

Answer:

Explanation:

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3 years ago

Burka [1]

Answer:

$\large \boxed{\text{E) 721 K; B) 86.7 g}}$

Explanation:

Question 7.

We can use the Combined Gas Laws to solve this question.

a) Data

p₁ = 1.88 atm; p₂ = 2.50 atm

V₁ = 285 mL; V₂ = 435 mL

T₁ = 355 K; T₂ = ?

b) Calculation

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{T_{1}}& =&\dfrac{p_{2}V_{2}}{T_{2}}\\\\\dfrac{1.88\times285}{355} &= &\dfrac{2.50\times 435}{T_{2}}\\\\1.509& = &\dfrac{1088}{T_{2}}\\\\1.509T_{2} & = & 1088\\T_{2} & = & \dfrac{1088}{1.509}\\\\ & = & \textbf{721K}\\\end{array}\\\text{The gas must be heated to $\large \boxed{\textbf{721 K}}$}$

Question 8. I

We can use the Ideal Gas Law to solve this question.

pV = nRT

n = m/M

pV = (m/M)RT = mRT/M

a) Data:

p = 4.58 atm

V = 13.0 L

R = 0.082 06 L·atm·K⁻¹mol⁻¹

T = 385 K

M = 46.01 g/mol

(b) Calculation

$\begin{array}{rcl}pV & = & \dfrac{mRT}{M}\\\\4.58 \times 13.0 & = & \dfrac{m\times 0.08206\times 385}{46.01}\\\\59.54 & = & 0.6867m\\m & = & \dfrac{59.54}{0.6867 }\\\\ & = & \textbf{86.7 g}\\\end{array}\\\text{The mass of NO$_{2}$ is $\large \boxed{\textbf{86.7 g}}$}$

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3 years ago