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3 years ago

Why does carbon have the ability to form multiple bonds?

1 answer:

7 0

Carbon has the ability to form multiple bonds because it has four valence electrons. Having four valence electrons means that carbon has a lot of space to form bonds with other atoms, or multiple bonds, in order to reach the full octet.

Hope this helps!! :)

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a glass container was initially charged with 1.50 mol of a gas sample at 3.75 atm and 21.7C. some of the gas was release as the

butalik [34]

Answer:

0.39 mol

Explanation:

Considering the ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L.atm/K.mol

At same volume, for two situations, the above equation can be written as:-

$\frac {{n_1}\times {T_1}}{P_1}=\frac {{n_2}\times {T_2}}{P_2}$

Given ,

n₁ = 1.50 mol

n₂ = ?

P₁ = 3.75 atm

P₂ = 0.998 atm

T₁ = 21.7 ºC

T₂ = 28.1 ºC

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (21.7 + 273.15) K = 294.85 K

T₂ = (28.1 + 273.15) K = 301.25 K

Using above equation as:

$\frac{{n_1}\times {T_1}}{P_1}=\frac{{n_2}\times {T_2}}{P_2}$

$\frac{{1.50\ mol}\times {294.85\ K }}{3.75\ atm}=\frac{{n_2}\times {301.25\ K }}{0.998\ atm}$

$n_2=\frac{{1.50}\times {294.85}\times 0.998}{3.75\times 301.25}\ mol$

Solving for n₂ , we get:

n₂ = 0.39 mol

7 0

3 years ago

The electron configuration of nitrogen (N)

erik [133]

Nitrogen atomic no.=7
electron=7its electronic configuration =1s2 2s2 2p3

5 0

4 years ago

Read 2 more answers

Can sombody pls help me with this

luda_lava [24]

Wouldn’t it be half of each? For 36 I guess is 18 and 54 will be 27, (NOT SURE)

6 0

3 years ago

A sample of HI (9.30×10^−3mol) was placed in an empty 2.00 L container at 1000 K. After equilibrium was reached, the concentrati

Sedaia [141]

Answer:

The answer is "29.081"

Explanation:

when the empty 2.00 L container of 1000 kg, a sample of HI (9.30 x 10-3 mol) has also been placed.

$\text{calculating the initial HI}= \frac{mol}{V}$

$=\frac{9.3 \times 10 ^ -3}{2}$

$=0.00465 \ Mol$

$\text{Similarly}\ \ I_2 \ \ \text{follows} \ \ H_2 = 0 }$

Its density of I 2 was 6.29x10-4 M if the balance had been obtained, then we have to get the intensity of equilibrium then:

$HI = 0.00465 - 2x\\\\ I_{2} \ eq = H_2 \ eq = 0 + x \\\\$

It is defined that:

$I_2 = 6.29 \times 10^{-4} \ M \\\\x = I_2 \\\\$

$HI \ eq= 0.00465 - 2x \\$

$=0.00465 -2 \times 6.29 \times 10^{-4} \\\\ = 0.00465 -\frac{25.16 }{10^4} \\\\ = 0.003392\ M$

Now, we calculate the position:

For the reaction $H 2(g) + I 2(g)\rightleftharpoons 2HI(g)$ , you can calculate the value of Kc at 1000 K.

data expression for Kc

$2HI \rightleftharpoons H_2 + I_2 \\\\\to Kc = \frac{H_2 \times I_2}{HI^2}$

$= \frac{6.29\times10^{-4} \times 6.29 \times 10^{-4}}{0.003392^2} \\\\= \frac{6.29\times 6.29 \times 10^{-8}}{0.003392^2} \\\\= \frac{39.564 \times 10^{-8}}{1.150 \times 10-5} \\\\= 0.034386$