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daser333 [38]
3 years ago
15

If a flask containing one liter of water was poured into 1,000 small cubes, then each cube would contain:

Chemistry
2 answers:
erastova [34]3 years ago
7 0
Every cube would contain 1 milliliter of water, or .01 liter
anzhelika [568]3 years ago
4 0

Answer is: 1 mL or 1 cm³.

V(water) = 1 L; volume of the water.

One liter is equal to 1000 milliliters (mL) or 1000 cubic centimetres (1000 cm³).

V(water) = 1 L · 1000 mL/L.

V(water) = 1000 mL.

V(cube) = 1000 mL ÷ 1000.

V(cube) = 1 mL; volume of the cube.

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The answer is (C) as the definitions states that atomic mass is the number of protons and neutrons.
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Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio
umka21 [38]

Answer : The value of K_p at this temperature is 66.7

Explanation : Given,

Pressure of PCl_3 at equilibrium = 0.348 atm

Pressure of Cl_2 at equilibrium = 0.441 atm

Pressure of PCl_5 at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

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

The expression of equilibrium constant K_p for the reaction will be:

K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}

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

K_p=\frac{(10.24)}{(0.348)(0.441)}

K_p=66.7

Therefore, the value of K_p at this temperature is 66.7

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It is easier to determine the electron configurations for the p-block elements in periods 1, 2, and 3 than to determine the elec
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Answer:

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  • Suppose a element be Ga .

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5 0
2 years ago
If 3.0 atm of pure HN3(g) is decomposed initially, what is the final total pressure in the reaction container? What are the part
lawyer [7]

This is an incomplete question, here is a complete question.

Hydrogen azide, HN₃, decomposes on heating by thefollowing unbalanced reaction:

HN_3(g)\rightarrow N_2(g)+H_2(g)

If 3.0 atm of pure HN₃ (g) is decomposed initially,what is the final total pressure in the reaction container? Whatare the partial pressures of nitrogen and hydrogen gas? Assume thatthe volume and temperature of the reaction container are constant.

Answer : The partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

Explanation :

The given unbalanced chemical reaction is:

HN_3(g)\rightarrow N_2(g)+H_2(g)

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen and nitrogen atoms are not balanced on both side of the reaction.

In order to balance the chemical equation, the coefficient '2' put before the HN_3 and the coefficient '3' put before the N_2 then we get the balanced chemical equation.

The balanced chemical reaction will be,

2HN_3(g)\rightarrow 3N_2(g)+H_2(g)

As we are given:

The pressure of pure HN_3 = 3.0 atm

p_{Total}=2\times p_{HN_3}=2\times 3.0atm=6.0atm

From the reaction we conclude that:

Number of moles of N_2 = 3 mol

Number of moles of H_2 = 1 mol

Now we have to calculate the mole fraction of N_2 and H_2

\text{Mole fraction of }N_2=\frac{\text{Moles of }N_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{3}{3+1}=0.75

and,

\text{Mole fraction of }H_2=\frac{\text{Moles of }H_2}{\text{Moles of }N_2+\text{Moles of }H_2}=\frac{1}{3+1}=0.25

Now we have to calculate the partial pressure of N_2 and H_2

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 6.0 atm

X_i = mole fraction of gas

p_{N_2}=X_{N_2}\times p_T

p_{N_2}=0.75\times 6.0atm=4.5atm

and,

p_{H_2}=X_{H_2}\times p_T

p_{H_2}=0.25\times 6.0atm=1.5atm

Thus, the partial pressure of N_2 and H_2 gases are, 4.5 atm and 1.5 atm respectively.

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3 years ago
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