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2 years ago

7

Please help with 14-16 Thank you!

1 answer:

Natalka [10]2 years ago

4 0

I don't know about 14, but 15 is (4), because a liquid draws in heat to turn into a gas. 16 is (2), because to turn into a cold solid, something has to release heat.

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Where will you find the smallest potential energy and kinetic energy when throwing a ball in the air?

Ne4ueva [31]

You will find the smallest when u release the ball and the highest when the ball is in mid air glad I could help will u please make me brainliest

8 0

2 years ago

How many liters of solution will contain 1.5 moles of CaCl2 if the solution is 6.0 M CaCl2?

sashaice [31]

Answer:

0.25

Explanation:

volume= number of moles over concentration

hence v=1.5/6

6 0

2 years ago

Carbon tetrachloride can be produced by the following reaction: Suppose 1.20 mol of and 3.60 mol of were placed in a 1.00-L flas

hjlf

The given question is incomplete. The complete question is :

Carbon tetrachloride can be produced by the following reaction:

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Suppose 1.20 mol $CS_2(g)$ of and 3.60 mol of $Cl_2(g)$ were placed in a 1.00-L flask at an unknown temperature. After equilibrium has been achieved, the mixture contains 0.72 mol of $CCl_4$ . Calculate equilibrium constant at the unknown temperature.

Answer: The equilibrium constant at unknown temperature is 0.36

Explanation:

Moles of $CS_2$ = 1.20 mole

Moles of $Cl_2$ = 3.60 mole

Volume of solution = 1.00 L

Initial concentration of $CS_2$ = $\frac{moles}{volume}=\frac{1.20mol}{1L}=1.20M$

Initial concentration of $Cl_2$ = $\frac{moles}{volume}=\frac{3.60mol}{1L}=3.60M$

The given balanced equilibrium reaction is,

$CS_2(g)+3Cl_2(g)\rightleftharpoons S_2Cl_2(g)+CCl_4(g)$

Initial conc. 1.20 M 3.60 M 0 0

At eqm. conc. (1.20-x) M (3.60-3x) M (x) M (x) M

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[S_2Cl_2]\times [CCl_4]}{[Cl_2]^3[CS_2]}$

Now put all the given values in this expression, we get :

$K_c=\frac{(x)\times (x)}{(3.60-3x)^3\times (1.20-x)}$

Given :Equilibrium concentration of $CCl_4$ , x = $\frac{moles}{volume}=\frac{0.72mol}{1L}=0.72M$

$K_c=\frac{(0.72)\times (0.72)}{(3.60-3\times 0.72)^3\times (1.20-0.72)}$

$K_c=0.36$

Thus equilibrium constant at unknown temperature is 0.36

4 0

2 years ago

Can someone please help me

mrs_skeptik [129]

The answer will be D

8 0

3 years ago

A 3.50 amp power supply is used to deposit chromium from a solution of CrCl3. How long will it take to deposit 100.0 grams of ch

Norma-Jean [14]

Explanation:

The given data is as follows.

Current (I) = 3.50 amp, Mass deposited = 100.0 g

Molar mass of Cr = 52 g

It is known that 1 faraday of electricity will deposit 1 mole of chromium. As 1 faraday means 96500 C and 1 mole of Cr means 52 g.

Therefore, 100 g of Cr will be deposited by "z" grams of electricity.

$z \times 52 g = 96500 \times 100 g$

z = $\frac{96500 \times 100 g}{52 g}$

= 185576.9 C

As we know that, Q = I × t

Hence, putting the given values into the above equation as follows.

Q = I × t

185576.9 C = $3.50 amp \times t$

t = 53021.9 sec

Thus, we can conclude that 100 g of Cr will be deposited in 53021.9 sec.

3 0

2 years ago