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egoroff_w [7]
3 years ago
8

For the following reaction, the equilibrium constant Kc is 2.0 at a certain temperature. The reaction is endothermic. What do yo

u expect to happen to the concentration of NO if the temperature is doubled? 2NOBr( g) ⇌ 2NO( g) + Br 2( g) The change in concentration of [NO] will depend on the size of the vessel. The concentration of NO will increase. The concentration of NO will decrease. A catalyst will be needed to make a change in concentration. There will be no change in [NO].
Chemistry
1 answer:
jeyben [28]3 years ago
7 0

Answer:

The correct answer is : 'the concatenation of NO will increase'.

Explanation:

Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

If the temperature is increased, so according to the Le-Chatlier's principle , the equilibrium will shift in the direction where increase in temperature occurs.

2NOBr(g)\rightleftharpoons 2NO(g) + Br_2(g)

As, this is an endothermic reaction, increasing temperature will add more heat to the system which move equilibrium in the forward reaction with decrease in temperature. Hence, the equilibrium will shift in the right direction.

So, the concatenation of NO will increase.

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A sample of helium (He) effuses 2.0 times faster than another gas. What is the molar mass of the other gas?
Mariana [72]

Answer:

The correct answer is 16 gram per mole.

Explanation:

Let A be the gas helium, and B be the unknown gas. It is clearly mentioned in the question that the effusion rate of helium gas is two times more than that of gas B. The molar mass of helium is 4.0 gram per mole. To solve the problem, Grahm's law is used, that is,  

Rate of effusion A/rate of effusion B = √ (Molar mass B/Molar mass of A

2.0 = √Molar mass of B/4.0 gram per mole.  

Now squaring both the sides we get,  

4.0 = Molar mass of B / 4.0

The molar mass of B = 16 gram per mole.  

6 0
3 years ago
If the pressure on a gas is increased, then its volume will...?
oksian1 [2.3K]

Decrease. I got you

4 0
3 years ago
How many sulfur atoms are in 0.45 mol BaSO4??
Lemur [1.5K]

(0.45 mol BaSO_{4})*(\frac{6.022*10^23 molecules}{mol})*(\frac{1 atom S}{1 molecule BaSO_{4}}) = 2.7*10^{23}

4 0
2 years ago
Read 2 more answers
What Kelvin temperature is the same as -13° Celsius?
zavuch27 [327]

Answer:

Degrees Kelvin = Degrees Celsius + 273.15

Degrees Kelvin = -13 + 273.15

= 260.15 °

Explanation:

8 0
3 years ago
Be sure to answer all parts. The percent by mass of bicarbonate (HCO3−) in a certain Alka-Seltzer product is 32.5 percent. Calcu
pochemuha

Answer : The volume of CO_2 will be, 514.11 ml

Explanation :

The balanced chemical reaction will be,

HCO_3^-+HCl\rightarrow Cl^-+H_2O+CO_2

First we have to calculate the  mass of HCO_3^- in tablet.

\text{Mass of }HCO_3^-\text{ in tablet}=32.5\% \times 3.79g=\frac{32.5}{100}\times 3.79g=1.23175g

Now we have to calculate the moles of HCO_3^-.

Molar mass of HCO_3^- = 1 + 12 + 3(16) = 61 g/mole

\text{Moles of }HCO_3^-=\frac{\text{Mass of }HCO_3^-}{\text{Molar mass of }HCO_3^-}=\frac{1.23175g}{61g/mole}=0.0202moles

Now we have to calculate the moles of CO_2.

From the balanced chemical reaction, we conclude that

As, 1 mole of HCO_3^- react to give 1 mole of CO_2

So, 0.0202 mole of HCO_3^- react to give 0.0202 mole of CO_2

The moles of CO_2 = 0.0202 mole

Now we have to calculate the volume of CO_2 by using ideal gas equation.

PV=nRT

where,

P = pressure of gas = 1.00 atm

V = volume of gas = ?

T = temperature of gas = 37^oC=273+37=310K

n = number of moles of gas = 0.0202 mole

R = gas constant = 0.0821 L.atm/mole.K

Now put all the given values in the ideal gas equation, we get :

(1.00atm)\times V=0.0202 mole\times (0.0821L.atm/mole.K)\times (310K)

V=0.51411L=514.11ml

Therefore, the volume of CO_2 will be, 514.11 ml

6 0
2 years ago
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