While in science class a student measured a cube. She

What is the pH of a 3.5 x 10 to the negative 11th M H+ solution

Ghella [55]

Answer:

Explanation:

-log(3.5 * 10^-11)

= 10.4559

Be careful how you put this into your calculator. I had to use Exp to get it to work properly.

-log

(3.5 * 10 exp -11)

=

4 0

3 years ago

Read 2 more answers

What are the sources of petrochemicals

Sindrei [870]

Answer:

Petrochemicals are chemical products derived from petroleum, although many of the same chemical compounds are also obtained from other fossil fuels such as coal and natural gas or from renewable sources such as corn, sugar cane, and other types of biomass.

4 0

3 years ago

Read 2 more answers

An equilibrium mixture of PCl 5 ( g ) , PCl 3 ( g ) , and Cl 2 ( g ) has partial pressures of 217.0 Torr, 13.2 Torr, and 13.2 To

antoniya [11.8K]

Answer: The new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

Explanation:

For the given chemical reaction:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of $K_p$ for above reaction follows:

$K_p=\frac{P_{PCl_5}}{P_{PCl_3}\times P_{Cl_2}}$ ........(1)

We are given:

$P_{PCl_5}=217.0torr$

$P_{PCl_3}=13.2torr$

$P_{Cl_2}=13.2torr$

Putting values in above equation, we get:

$K_p=\frac{217.0}{13.2\times 13.2}\\\\K_p=1.24$

Now we have to calculate the new partial pressure of $Cl_2$ .

$P_{PCl_5}+P_{PCl_3}+P_{Cl_2}=P_{Total}$

$217.0torr+13.2torr+P_{Cl_2}=263.0torr$

$P_{Cl_2}=32.8torr$

The reaction is re-established and proceed to right direction by Le-Chatelier's principle to cancel the effect of addition of $Cl_2$ .

Now, the equilibrium is shifting to the reactant side. The equation follows:

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initial: 13.2 32.8 217.0

At eqm: 13.2-x 32.8-x 217.0+x

Putting values in expression 1, we get:

$1.24=\frac{(217.0+x)}{(13.2-x)(32.8-x)}\\\\x=40.4,6.38$

Neglecting the 40.4 value of 'x' because pressure can not be more than initial partial pressure.

Thus, the value of 'x' will be, 6.38 torr.

Now we have to calculate the new partial pressures after equilibrium is reestablished.

Partial pressure of $PCl_5$ = (217.0+x) = (217.0+6.38) = 223.4 torr

Partial pressure of $PCl_3$ = (13.2-x) = (13.2-6.38) = 6.82 torr

Partial pressure of $Cl_2$ = (32.8-x) = (32.8-6.38) = 26.4 torr

Hence, the new partial pressures of $PCl_5,PCl_3\text{ and }Cl_2$ when equilibrium is re-established are 223.4 torr, 6.82 torr and 26.4 torr respectively.

7 0

3 years ago

20. Given the reaction at equilibrium:

Gemiola [76]

PV = nRT

If pressure increases, so will moles.

The answer is 2) Increase.

8 0

2 years ago

Which of the following conditions on Mars would be the first to kill a human who is unprotected and unassisted by life support?

TiliK225 [7]

Answer:

D

Excess solar radiation due to a missing magnetic field.

Explanation: Solar proton events (SPEs) are bursts of energetic protons accelerated by the Sun. They occur relatively rarely and can produce extremely high radiation levels. Without thick shielding, SPEs are sufficiently strong to cause acute radiation poisoning and death.

Hope this hels

plz mark brainliest

6 0

3 years ago

Read 2 more answers