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3 years ago

8

The half-life of radioactive element krypton-91 is 10 seconds. if 16 grams of krypton-91 are initially present, how many grams a

re present after 10 seconds? after 50 seconds?

1 answer:

earnstyle [38]3 years ago

6 0

10 seconds = 8grams
then just divide by 2 another 4 times...
= 0.5grams after 50 seconds

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At 27.00 °c a gas has a volume of 6.00 L . What will the volume be at 150.0 °c

mojhsa [17]

T₁ = 27°C = 27 + 273 = 300K, V₁ = 6 L,

T₂ = 150°C = 150 + 273 = 423K, V₂ = ?,

By Charles' Law: V₁/T₁= V₂/T₂

6/300 = V₂/423

423*(6/300) = V₂

8.46 = V₂

Volume at 150°C =8.46 L.

3 0

3 years ago

When the temperature of a gas changes, it's volume decreases from 12 cm3 to 7 cm3 if the final temperature is measured to be 18°

ElenaW [278]

Answer:

The initial temperature is 499 K

Explanation:

Step 1: Data given

initial volume = 12 cm3 = 12 mL

Final volume = 7 cm3 = 7mL

The final temperature = 18 °C = 291 K

Step 2: Calculate the initial temperature

V1/T1 = V2/T2

⇒with V1 = the initial volume = 0.012 L

⇒with T1 = the initial volume = ?

⇒with V2 = the final volume 0.007 L

⇒with T2 = The final temperature = 291 K

0.012 / T1 = 0.007 / 291

0.012/T1 = 2.4055*10^-5

T1 = 0.012/2.4055*10^-5

T1 = 499 K

The initial temperature is 499 K

8 0

3 years ago

A container has a volume of 2.79 L and a pressure of 5.97 atm. If the pressure changes to 1460 mm Hg, what is the container’s ne

maw [93]

Answer:

8.68 L is the new volume

Explanation:

You use Boyle's law for this.

$P_{1}V_{1}=P_{2}V_{2}$

$P_{1}$ = first pressure

$P_{2}$ = second pressure

$V_{1}$ = first volume

$V_{2}$ = second volume

Convert pressure from atm to mmHg (use same units):

5.97 x 760 = 4537.2 -> 4.54 x 10³

...maintain 3 significant figures in calculation, and round as needed...

(4.54 x 10³ mmHg)(2.79 L) = (1460 mmHg)( $V_{2}$ )

(4.54 x 10³ mmHg)(2.79 L) / (1460 mmHg) = $V_{2}$ = 8.68 L

Hope this helps :)

5 0

3 years ago

givi [52]

Ummm I just need to answer questions sorry!!!

8 0

3 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

3 years ago