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3 years ago

PLEASE ANSWER BOTH QUESTIONS I WILL MARK YOU BRAINLIEST IF IT IS CORRECT!!

Chemistry

1 answer:

seropon [69]3 years ago

7 0

Answer:

1st Question: A

2nd Question: B

Explanation:

The 1st answer would be A because if a sample is at absolute zero then the sample is at its lowest temperature none of the molecules would be able to move, this is because lower temperature= lower kinetic energy.

The 2nd answer would be B because if a sample has more temperature it speeds up it has more temperature and more kinetic energy, meaning it would move faster because there is more temperature.

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2A (g) + Y (g) <-- --> 3C (g) + D (g)

insens350 [35]

Answer:

A. K = 59.5

Explanation:

Hello there!

In this case, since this reaction seems to start moving leftwards due to the fact that neither A nor Y are present at equilibrium, we should rewrite the equation:

3C (g) + D (g) <-- --> 2A (g) + Y (g)

Thus, the equilibrium expression is:

$K^{left}=\frac{[A]^2[Y]}{[C]^3[D]}$

Next, according to an ICE table for this reaction, we find that:

$[A]=2x$

$[Y]=x$

$[C]=0.651M-3x$

$[D]=0.754M-x$

Whereas x is calculated by knowing that the [C] at equilibrium is 0.456M; thus:

$x=\frac{0.651-0.456}{3} =0.065M$

Next, we compute the rest of the concentrations:

$[A]=2(0.065M)=0.13M$

$[Y]=0.065M$

$[D]=0.754M-0.065M=0.689M$

Thus, the equilibrium constant for the leftwards reaction is:

$K^{left}=\frac{(0.13M)^2(0.065M)}{(0.456M)^3(0.689M)}=0.0168$

Nonetheless, we need the equilibrium reaction for the rightwards reaction; thus, we take the inverse to get:

$K^{right}=\frac{1}{0.0168}=59.5$

Therefore, the answer would be A. K = 59.5.

Regards!

6 0

3 years ago

A chlorine Cl and bromine Br atom are adsorbed on a small patch of surface (see sketch at right). This patch is known to contain

MrRissso [65]

Explanation:

It is given that possible number of ways the Cl and Br can be absorbed initially are 100.

S, possible number of ways by which Br can be desorbed is as follows.

$100 \times 99$

Now, we will calculate the change in entropy as follows.

$\Delta S = k_{B} ln (\frac{W}{W_{o}})$

where, $k_{B}$ = Boltzmann constant = $1.38 \times 10^{-23}$

$\Delta S$ = change in entropy

Therefore, we will calculate the change in entropy as follows.

$\Delta S = k_{B} ln (\frac{W}{W_{o}})$

= $1.38 \times 10^{-23} J/K \times ln (\frac{100}{100 \times 99})$

= $1.38 \times 10^{-23} J/K \times -4.595$

= $-6.34 \times 10^{-23} J/K$

Thus, we can conclude that the change in entropy is $-6.34 \times 10^{-23} J/K$ .

5 0

3 years ago

When 1 mol CS2(l) forms from its elements at 1 atm and 25°C, 89.7 kJ of heat is absorbed, and it takes 27.7 kJ to vaporize 1 mol

Kitty [74]

Answer:

There is 117.4 kJ of heat absorbed

Explanation:

<u>Step 1: </u>Data given

Number of moles CS2 = 1 mol

Temperature = 25° = 273 +25 = 298 Kelvin

Heat absorbed = 89.7 kJ

It takes 27.7 kJ to vaporize 1 mol of the liquid

<u>Step 2:</u> Calculate the heat that is absorbed

C(s) + 2S(s) → CS2(l) ΔH = 89.7 kJ (positive since heat is absorbed)

CS2(l) → CS2(g) ΔH = 27.7 kJ (positive since heat is absorbed)

We should balance the equations, before summing, but since they are already balanced, we don't have to change anything.

C(s) + 2S(s)---> CS2 (g)

ΔH = 89.7 + 27.7 = 117.4 kJ

There is 117.4 kJ of heat absorbed

6 0

3 years ago

I need helpppp will pay anything !!

nydimaria [60]

Answer: iss as kfnandn

Explanation:

7 0

3 years ago

Use the ideal gas law to calculate the pressure in atmospheres of 0.21 mol of helium (He) at 16°C & occupying 2.53 L. You mu

chubhunter [2.5K]

Answer:

The answer to your question is 2.32 atm

Explanation:

Data

P = ?

n = 0.214

V = 2.53 L

T = 61°C

R = 0.082 atm L/mol°K

Formula

PV = nTR

solve for P

P = nRT/V

Process

1.- Calculate the temperature in K

°K = °C + 273

°K = 61 + 273

= 334

2.- Substitution

P = (0.214 x 0.082 x 334) / 2.53

3.- Simplification

P = 5.86/2.53

4.- Result

P = 2.32 atm

4 0

3 years ago