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2 years ago

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What volume of 0.436 M barium nitrate solution is needed to prepare 240.0 mL of a barium nitrate solution that is 0.300 M in nit

rate ion?
a. 349 mL

b. 82.6 mL

c. 55.0 mL

d. 165 mL

e. 330. mL

1 answer:

Advocard [28]2 years ago

7 0

The correct answer would be A

Explanation: I did the math

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What is the mass of one mole of hydrogen molecules?

Lilit [14]

Answer:

2 grams

Explanation:

One MOLE of hydrogen atoms contains the same number of atoms as the number of hydrogen molecules in one MOLE of hydrogen molecules, i.e., Avagadros number. However, one mole of hydrogen atoms has a mass of 1 gram while one MOLE of hydrogen molecules has a mass of 2 grams.

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2 years ago

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For the reaction: N2(g) + 2 O2(g) ⇌ 2 NO2(g), Kc = 8.3 × 10-10 at 25°C. What is the concentration of N2 gas at equilibrium when

KATRIN_1 [288]

Answer: Concentration of N₂ is 4.8. $10^{9}$ M.

Explanation: $K_{c}$ is a constant of equilibrium and it is dependent of the concentrations of the reactants and the products of a balanced reaction. For

N2(g) + 2 O2(g) ⇄ 2 NO2(g)

$K_{c}$ = $\frac{[NO2]^{2} }{[N2][O2]^{2} }$

From the question concentration of NO2 is twice of O2:

[NO2] = 2[O2]

Substituting this into $K_{c}$ :

$K_{c}$ = $\frac{[2O2]^{2} }{[N2][O2]^{2} }$

8.3. $10^{-10}$ = $\frac{4O2^{2} }{[N2].O2^{2} }$

[N2] = $\frac{4O2^{2} }{8.3.10^{-10}.O2^{2} }$

[N2] = $\frac{4}{8.3.10^{-10} }$

[N2] = 4.8. $10^{9}$

The concentration of N2 in the equilibrium is [N2] = 4.8. $10^{9}$ M.

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2 years ago

Change 32L of carbon dioxide into moles of carbon dioxide

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32L —> 32000g —> 727.116 Moles (rounded)

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2 years ago

Which phase change temperatures are identical on the phase diagram? Select all that apply.

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boiling point - condensation point

is the answer i would choose because it makes more scene

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2 years ago

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The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Sati [7]

Answer:

a. 113 min

Explanation:

Considering the equilibrium:-

2N₂O₅ ⇔ 4NO₂ + O₂

At t = 0 125 kPa

At t = teq 125 - 2x 4x x

Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x

125 - 3x = 176 kPa

x = 17 kPa

Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $2.8\times 10^{-3}$ min⁻¹

Initial concentration $[A_0]$ = 125 kPa

Final concentration $[A_t]$ = 91 kPa

Time = ?

Applying in the above equation, we get that:-

$91=125e^{-2.8\times 10^{-3}\times t}$

$125e^{-2.8\times \:10^{-3}t}=91$

$-2.8\times \:10^{-3}t=\ln \left(\frac{91}{125}\right)$

$t=113\ min$

3 0

2 years ago