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vaieri [72.5K]
3 years ago
14

When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilibr

ium, the concentration of N2O4 is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction below at
400 K? N2O4(g) ⇌ 2 NO2(g)
Chemistry
1 answer:
Amanda [17]3 years ago
6 0

<u>Answer:</u> The value of K_c for the given reaction is 1.435

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}

Given mass of N_2O_4 = 9.2 g

Molar mass of N_2O_4 = 92 g/mol

Volume of solution = 0.50 L

Putting values in above equation, we get:

\text{Molarity of solution}=\frac{9.2g}{92g/mol\times 0.50L}\\\\\text{Molarity of solution}=0.20M

For the given chemical equation:

                 N_2O_4(g)\rightleftharpoons 2NO_2(g)

<u>Initial:</u>          0.20

<u>At eqllm:</u>     0.20-x        2x

We are given:

Equilibrium concentration of N_2O_4 = 0.057

Evaluating the value of 'x'

\Rightarrow (0.20-x)=0.057\\\\\Rightarrow x=0.143

The expression of K_c for above equation follows:

K_c=\frac{[NO_2]^2}{[N_2O_4]}

[NO_2]_{eq}=2x=(2\times 0.143)=0.286M

[N_2O_4]_{eq}=0.057M

Putting values in above expression, we get:

K_c=\frac{(0.286)^2}{0.143}\\\\K_c=1.435

Hence, the value of K_c for the given reaction is 1.435

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If 10.0 mL of a .600 M of HNO3 reacts with 31.0 mL of .700M Ba(OH)2 solution, what is the molarity of Ba(OH)2 after the reaction
Tasya [4]

Answer:

<u></u>

  • <u>0.456M</u>

Explanation:

<u>1. Balanced molecular equation</u>

     2HNO_3+Ba(OH)_2\rightarrow Ba(NO_3)_2+2H_2O

<u>2. Mole ratio</u>

     \dfrac{2molHNO_3}{1molBa(OH)_2}

<u>3. Moles of HNO₃</u>

  • Number of moles = Molarity × Volume in liters
  • n = 0.600M × 0.0100 liter = 0.00600 mol HNO₃

<u>4. Moles Ba(OH)₂</u>

  • n = 0.700M × 0.0310 liter = 0.0217 mol

<u>5. Limiting reactant</u>

Actual ratio:

   \dfrac{0.0600molHNO_3}{0.0217molBa(OH)_2}\approx0.28

Since the ratio of the moles of HNO₃ available to the moles of Ba(OH)₂ available is less than the theoretical mole ratio, HNO₃ is the limiting reactant.

Thus, 0.006 moles of HNO₃ will react completely with 0.003 moles of Ba(OH)₂ and 0.0217 - 0.003 = 0.0187 moles will be left over.

<u>6. Final molarity of Ba(OH)₂</u>

  • Molarity = number of moles / volume in liters
  • Molarity = 0.0187 mol / (0.0100 + 0.0031) liter = 0.456M
5 0
3 years ago
Given any two elements within a group, is the element with the larger atomic number likely to have a larger or smaller atomic ra
zhenek [66]

Answer:

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

Explanation:

Atomic radii trend along group:

As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.

Consider the example of sodium and potassium.

Sodium is present above the potassium with in same group i.e, group one.

The atomic number of sodium is 11 and potassium 19.

So potassium will have larger atomic radius as compared to sodium.

Atomic radius of sodium = 227 pm

Atomic radius of potassium = 280 pm

3 0
3 years ago
Which best describes the field of chemistry?
qaws [65]

Answer:

Fundamentally, chemistry is the study of matter and change. The way that chemists study matter and change and the types of systems that are studied varies dramatically. Traditionally, chemistry has been broken into five main subdisciplines: Organic, Analytical, Physical, Inorganic, and Biochemistry.

hope this helps :)

5 0
3 years ago
As part of your job you are asked to make 1 liter of a 0.5 molar sucrose solution. how much sucrose (c12h22o11) do you need? use
kakasveta [241]

Answer:-  171 g

Solution:- It asks to calculate the grams of sucrose required to make 1 L of 0.5 Molar solution of it.

We know that molarity is moles of solute per liter of solution.

If molarity and volume is given then, moles of solute is molarity times volume in liters.

moles of solute = molarity* liters of solution

moles of solute = 0.5*1 = 0.5 moles

To convert the moles to grams we multiply the moles by molar mass.

Molar mass of sucrose = 12(12) + 22(1) + 11(16)  

= 144 + 22 + 176

= 342 grams per mol

grams of sucrose required = moles * molar mass

grams of sucrose required = 0.5*342  = 171 g

So, 171 g of sucrose are required to make 1 L of 0.5 molar solution.




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