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Arte-miy333 [17]

3 years ago

8

Can someone please check my answer?

2 answers:

goblinko [34]3 years ago

5 0

yes correctly ☺️☺️

Explanation:

thank you ❣️

77julia77 [94]3 years ago

4 0

Answer:

yes your answer is correct

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If the volume of a container of air is reduced by one-half what happens to the partial pressure of oxygen within the container?

katovenus [111]

Decreasing Pressure. The combined gas law states that the pressure of a gas is inversely related to the volume and directly related to the temperature. If temperature is held constant, the equation is reduced to Boyle's law. <em>Therefore, if you decrease the pressure of a fixed amount of gas, its volume will increase.</em>

<em>thanxx</em>

<em>moderator will see it i bett and will delete itt</em>

5 0

4 years ago

Read 2 more answers

Which of the following might be in for a rough orbit around the Sun because of debris in their paths? (Choose all that apply.)

GuDViN [60]

Answer:

cerus neptune and eris

Explanation:

6 0

3 years ago

Plz help, I will give brainliest.

baherus [9]

Answer: The value of $K_c$ is 2

Explanation:

Moles of $N_2O_4$ = 1.0 mole

Volume of solution = 1.00 L

Initial concentration of $N_2O_4$ = $\frac{1.0mol}{1.00L}=1.0M$

Equilibrium concentration of $NO_2$ = $\frac{1.0mol}{1.00L}=1.0M$

The given balanced equilibrium reaction is,

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

Initial conc. 1.0 M 0 M

At eqm. conc. (1.0-x) M (2x) M

The expression for equilibrium constant for this reaction will be,

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

Given : 2x = 1.0

x= 0.5

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

$K_c=\frac{(1.0)^2}{(1.0-0.5)}$

$K_c=2$

Thus the value of $K_c$ is 2

4 0

3 years ago

Two friends are arguing about the structure of the universe. Caleb says that the universe is made of many galaxies that are

Nadusha1986 [10]

Answer:

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Explanation:

4 0

3 years ago

2.61 kilograms of water in a container have a pressure of 200 kPa and temperature of 200°C . What is the volume of this containe

Arisa [49]

<u>Answer:</u> The volume of the container is $2.8497m^3$

<u>Explanation:</u>

To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 200 kPa

V = volume of container = ? L

m = Given mass of water = 2.61 kg = 2610 g (Conversion factor: 1kg = 1000 g)

M = Molar mass of water = 18 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $200^oC=[200+273]K=473K$

Putting values in above equation, we get:

$200kPa\times V=\frac{2610g}{18g/mol}\times 8.31\text{L kPa }\times 473K\\\\V=2849.7L$

Converting this into cubic meter, we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow \frac{1m^3}{1000L}\times 2849.7L$

$\Rightarrow 2.8497m^3$

Hence, the volume of the container is $2.8497m^3$

6 0

3 years ago