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Romashka-Z-Leto [24]
3 years ago
7

Which of the following changes would have no effect on the equilibrium position of the reaction below? 2 NOBR (g) 2 NO (g)+ Br2

(g) A) adding NOBr. B) removing NO C) doubling the concentration of NO at the same time as the concentration of Br2 is cut in half. D) cutting the concentrations of both NOBr and NO in half
Chemistry
1 answer:
nasty-shy [4]3 years ago
5 0

Answer:

D) cutting the concentrations of both NOBr and NO in half

Explanation:

The equilibrium reaction given in the question is as follows -

2NOBr ( g ) ↔  2NO ( g ) + Br₂ ( g )

The equilibrium constant for the above reaction can be written as -

K = [ NO ]² [ Br₂ ] / [ NOBr ] ²

Therefore from the condition given in the question , the changes that will not affect the equilibrium will be , reducing the concentration of both NOBr and NO to half ,

Hence ,

the new concentrations are as follows -

[ NoBr ] ' = 1/2 [ NoBr ]

[ NO ] ' = 1/2  [ NO ]

Hence the new equilibrium constant equation can be written as -

K ' = [ NO ] ' ² [ Br₂ ] / [ NOBr ]  ' ²

Substituting the new concentration terms ,

K ' = 1/2  [ NO ] ² [ Br₂ ] / 1/2 [ NoBr ]  ²

K ' = 1/4  [ NO ] ² [ Br₂ ] / 1/4 [ NoBr ]  ²

The value of 1 / 4 in the numerator and the denominator is cancelled -

K ' =   [ NO ] ² [ Br₂ ] /  [ NoBr ]  ²

Hence ,

K' = K

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If the chemist mistakenly makes 250 mL of solution instead of the 200 mL, what molar concentration of sodium nitrate will the ch
Evgesh-ka [11]

Answer:

0.120M is the concentration of the solution

Explanation:

<em>Assuming the mass of sodium nitrate dissolved was 2.552g</em>

<em />

Molar concentration is an unit of concentration widely used in chemsitry defined as the moles of solute (In this case NaNO3) in 1L of solution.

To find this question we must find the moles of NaNO3 in 2.552g. With this mass and the volume (250mL = 0.250L) we can find molar concentration as follows:

<em>Moles NaNO3 -Molar mass: 84.99g/mol-</em>

2.552g * (1mol / 84.99g) = 0.0300 moles NaNO3

<em>Molar concentration:</em>

0.0300 moles NaNO3 / 0.250L =

<h3>0.120M is the concentration of the solution</h3>
7 0
3 years ago
Calculate the standard entropy of vaporization of ethanol at its boiling point, 352 K. The standard molar enthalpy of vaporizati
Korvikt [17]

Answer : The correct option is, (b) +115 J/mol.K

Explanation :

Formula used :

\Delta S=\frac{\Delta H_{vap}}{T_b}

where,

\Delta S = change in entropy

\Delta H_{vap} = change in enthalpy of vaporization = 40.5 kJ/mol

T_b = boiling point temperature = 352 K

Now put all the given values in the above formula, we get:

\Delta S=\frac{\Delta H_{vap}}{T_b}

\Delta S=\frac{40.5kJ/mol}{352K}

\Delta S=115J/mol.K

Therefore, the standard entropy of vaporization of ethanol at its boiling point is +115 J/mol.K

8 0
3 years ago
A binary compound created by reaction of bismuth and an unknown element E contains 52.07% Bi and 47.93% E by mass. If the formul
yulyashka [42]

Answer:

Atomic mass of E is 128.24

Explanation:

  • The percentage composition by mass of an element in a compound is given by dividing the mass of the element by the total mass of the compound and expressing it as a percentage.
  • In this case; the compound Bi₂E₃

Percentage composition of bismuth = 52.07%

Percentage composition of E = 47.93%

Mass Bismuth in the compound is (2×208.9804) = 417.96 g

Therefore,

To calculate the atomic mass of E

52.07% = 417.96 g

47.93% = ?

            = (47.93 × 417.96 ) ÷ 52.07 %

            = 384.729

         E₃ = 384.729

Therefore; E = 384.729 ÷ 3

                     = 128.24  

The atomic mass of E is 128.24

6 0
2 years ago
Which of the following is the best example of potential energy?
strojnjashka [21]
Answer :

A car stopped at the top of a hill

Explanation :

Definition of potential energy : energy stored that depends upon the relative position of various parts of a system
5 0
2 years ago
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
3 years ago
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