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alexdok [17]
3 years ago
13

A student uses 0.0821 L• atm/mol • K as the value of the gas constant. What is most likely true about the variables in the ideal

gas law?
P has the units of liters • atmospheres, and T has the units of kelvin.

P has the units of liters • atmospheres, and T has the units of degrees Celsius.

V has the units of liters, and T has the units of kelvin.

V has the units of liters, and T has the units of degrees Celsius.
Chemistry
2 answers:
Ray Of Light [21]3 years ago
7 0
Considering P represents pressure why god would it be measured in liters. Ignore the first 2 choices.
With all gas law related questions you never use Celsius temperatures.
The answer has to be V has units of liters, and T has the units of kelvin.
serious [3.7K]3 years ago
4 0

Answer is: V has the units of liters, and T has the units of kelvin.

Ideal gas law: p·V = n·R·T.  

atm · L = mol · L·atm/mol·K · K; both side of equatation have same values.

R = 0,08206 L·atm/mol·K; universal gas constant.  

p is pressure of the gas, unit is standard atmosphere (atm).

V is volume of the gas, unit is liters (L).

n is amount of substance of the gas; unit is mole (mol).

T is temperature of the gas, unit is Kelvin (K).

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Why does liquid stay at a constant temperature while it is boiling?
GaryK [48]
It is because say water boils at 212 F, if it goes higher at 213 it would get so much heat and energy it turns into a gas, so it cannot stay a liquid with 213 because at that point it would be gas, thus when water reaches 212 it's max if it goes any higher it will be gas
4 0
3 years ago
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Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:
posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of HNO_3 = \frac{15.5g}{63g/mol\times 9.5L}=0.026M

Equilibrium concentration of NO = \frac{16.6g}{30g/mol\times 9.5L}=0.058M

Equilibrium concentration of NO_2 = \frac{22.5g}{46g/mol\times 9.5L}=0.051M

Equilibrium concentration of H_2O = \frac{189.0g}{18g/mol\times 9.5L}=1.10M

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c  

For the given chemical reaction:

2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)

The expression for K_c is written as:

K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}

K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}

K_c=3.72

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

5 0
3 years ago
I need help with this question
Leviafan [203]

Answer:

i do not know the answer but pls give a heart and five starts so i can ask questions pls

Explanation:

3 0
3 years ago
The half life of 226/88 Ra is 1620 years. How much of a 12 g sample of 226/88 Ra will be left after 8 half lives?
Aloiza [94]

Answer:

0.0468 g.

Explanation:

  • The decay of radioactive elements obeys first-order kinetics.
  • For a first-order reaction: k = ln2/(t1/2) = 0.693/(t1/2).

Where, k is the rate constant of the reaction.

t1/2 is the half-life time of the reaction (t1/2 = 1620 years).

∴ k = ln2/(t1/2) = 0.693/(1620 years) = 4.28 x 10⁻⁴ year⁻¹.

  • For first-order reaction: <em>kt = lna/(a-x).</em>

where, k is the rate constant of the reaction (k = 4.28 x 10⁻⁴ year⁻¹).

t is the time of the reaction (t = t1/2 x 8 = 1620 years x 8 = 12960 year).

a is the initial concentration (a = 12.0 g).

(a-x) is the remaining concentration.

∴ kt = lna/(a-x)

(4.28 x 10⁻⁴ year⁻¹)(12960 year) = ln(12)/(a-x).

5.54688 = ln(12)/(a-x).

Taking e for the both sides:

256.34 = (12)/(a-x).

<em>∴ (a-x) = 12/256.34 = 0.0468 g.</em>

8 0
3 years ago
Help not to sure what its asking I thought it was 2 but there ain't a 2 ​
Shtirlitz [24]

the answer is d. this is due to the fact a proton weighs 2000 times more then a electron

5 0
3 years ago
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