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4 years ago

Solids can often diffuse into a liquid, but a solid cannot diffuse into another solid. true false

Chemistry

2 answers:

STatiana [176]4 years ago

7 0

Its False

hope that helps

Burka [1]4 years ago

4 0

Answer:

True.

Explanation:

As the amount of temperature is directly measured by the energy contained in the substance. When the average kinetic energy increases the molecules of solids then diffuse into liquid waters and then form gas with the increase in this energy.

But to diffuse into another solid the matter must have a surface that allows for the mixing to occur at higher temperature and pressure, which is mostly seen in liquids and gases.

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Calculate the pressure of 2 mol of a gas at 300 K in 8 L container.

Sergio039 [100]

Answer:

4800

Explanation:

using my Cal ex to solve the question

calculation goes like this

2*300*8=4800

3 0

3 years ago

What is the molar mass of sulfur trioxide?

tatyana61 [14]

Answer:

80.07 g/mol

Explanation:

Sulfur's g/mol = 32.07

Oxygen's g/mol = 16.00

32.07 + 16.00(3) = 80.07 g/mol

6 0

3 years ago

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Consider the following reaction. N2(g) + O2(g) = 2 NO(g) (a) If Kc for this reaction is 1.5 ✕ 10−10, and a reaction system at eq

svet-max [94.6K]

Answer:

For a: The concentration of NO at equilibrium is $5.27\times 10^{-7}M$

For b: The value of $K_c'$ is $6.66\times 10^{10}$

For c: The correct answer is product favored.

Explanation:

For a:

The given chemical reaction follows:

$N_2(g)+O_2(g)\rightleftharpoons 2NO(g)$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO]^2}{[N_2][O_2]}$

We are given:

$K_c=1.5\times 10^{-10}$

$[N_2]_{eq}=0.043M$

$[O_2]_{eq}=0.043M$

Putting values in above equation, we get:

$1.5\times 10^{-10}=\frac{[NO]^2}{0.043\times 0.043}$

$[NO]=\sqrt{(1.5\times 10^{-15}\times 0.043\times 0.043)}=5.27\times 10^{-7}M$

Hence, the concentration of NO at equilibrium is $5.27\times 10^{-7}M$

For b:

The given chemical reaction follows:

$2NO(g)\rightleftharpoons N_2(g)+O_2(g)$

The expression of $K_c'$ for above equation follows:

$K_c'=\frac{[N_2][O_2]}{[NO]^2}$

As, the above reaction is the reverse of equation in part a. So, the value of $K_c'$ will be inverse of $K_c$

$K_c'=\frac{1}{K_c}\\\\K_c'=\frac{1}{1.5\times 10^{-10}}=6.66\times 10^{10}$

Hence, the value of $K_c'$ is $6.66\times 10^{10}$

For c:

There are 3 conditions:

When $K_{c}>1$ ; the reaction is product favored.
When $K_{c}$ ; the reaction is reactant favored.
When $K_{c}=1$ ; the reaction is in equilibrium.

For the reaction in part 'b', the value of $K_c'$ is $6.66\times 10^{10}$

The value of $K_c'$ is very high than 1. So, the equilibrium in part 'b' is product favored.

Hence, the correct answer is product favored.

7 0

3 years ago

You added 5.00 ml of 0.40 m naoh in methanol to 20.00 ml of cooking oil. Calculate the number of moles of vegetable oil

goldenfox [79]

Answer:

$n_{oil}=0.002mol$

Explanation:

Hello,

In this case, considering that the sodium hydroxide reacts with the cooking oil, the following relationship is applied:

$n_{oil}=n_{NaOH}\\n_{oil}=M_{NaOH}*V_{NaOH}\\n_{oil}=0.4\frac{mol}{L} * 0.005L\\n_{oil}=0.002mol$

Best regards.

4 0

3 years ago

Any girl to talk to me?

Artist 52 [7]

Answer:

Sorry sir/ma'am brainly is not for this stuff.

Explanation:

5 0

3 years ago

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