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3 years ago

How is all matter alike?

Chemistry

2 answers:

Alex_Xolod [135]3 years ago

5 0

Definite shape and definite volume makes them all alike

nignag [31]3 years ago

4 0

All matter (solid, liquid or gas) has a definite shape and occupies a specific volume.

<h2>Hope it helps...</h2>

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When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

Amanda [17]

<u>Answer:</u> The value of $K_c$ for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

Given mass of $N_2O_4$ = 9.2 g

Molar mass of $N_2O_4$ = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M$

For the given chemical equation:

$N_2O_4(g)\rightleftharpoons 2NO_2(g)$

<u>Initial:</u> 0.20

<u>At eqllm:</u> 0.20-x 2x

We are given:

Equilibrium concentration of $N_2O_4$ = 0.057

Evaluating the value of 'x'

$\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143$

The expression of $K_c$ for above equation follows:

$K_c=\frac{[NO_2]^2}{[N_2O_4]}$

$[NO_2]_{eq}=2x=(2\times 0.143)=0.286M$

$[N_2O_4]_{eq}=0.057M$

Putting values in above expression, we get:

$K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435$

Hence, the value of $K_c$ for the given reaction is 1.435

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3 years ago

What does combustion mean

frutty [35]

Answer:

combustion is a high-temperature exothermic redox chemical reaction between a fuel and an oxidant, usually atmospheric oxygen, that produces oxidized, often gaseous products, in a mixture termed as smoke.

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2 years ago

How many moles is 22.4 liters of oxygen gas at standard temperature and pressure represent

makkiz [27]

Answer:

So 1 mole

Explanation:

PV = nRT

P = Pressure atm

V = Volume L

n = Moles

R = 0.08206 L·atm·mol−1·K−1.

T = Temperature K

standard temperature = 273K

standard pressure = 1 atm

22.4 liters of oxygen

Ok so we have

V = 22.4

P = 1 atm

PV = nRT

n = PV/RT

n = 22.4/(0.08206 x 273)

n = 22.4/22.40

n = 1 mole

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