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Lelechka [254]
3 years ago
12

How many molecules of CO2 are contained in 4.40g of CO2

Chemistry
1 answer:
Alexus [3.1K]3 years ago
5 0

Answer:

6.02 x 10²²molecules

Explanation:

Given parameters:

Mass of CO₂  = 4.4g

Unknown:

Number of molecules in CO₂  = ?

Solution:

To find the number of molecules in the given mass, we have to find the number of moles in the compound first;

   Number of moles  = \frac{mass}{molar mass}  

Molar mass of CO₂  = 12 + 2(16) = 44g/mol

Insert the parameters and solve;

   Number of moles  = \frac{4.4g}{44g/mol}   =  0.1mol

  1 mole of a substance contains 6.02 x 10²³molecules

  0.1 mole of CO₂ will contain 0.1 x 6.02 x 10²³molecules

                                                 = 6.02 x 10²²molecules

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2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea
Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}

Moles of HI = 0.550 moles

Volume of container = 2.00 L

\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M

For the given chemical equation:

                          2HI(g)\rightleftharpoons H_2(g)+I_2(g)

<u>Initial:</u>                  0.275

<u>At eqllm:</u>           0.275-2x      x         x

The expression of K_c for above equation follows:

K_c=\frac{[H_2][I_2]}{[HI]^2}

We are given:

K_c=0.0156

Putting values in above expression, we get:

0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275

Neglecting the negative value of 'x' because concentration cannot be negative

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Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

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Answer:

1.33 Å

Explanation:

Given that the edge length , a of the KCl which forms the FCC lattice = 6.28 Å

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For the FCC lattice in which the anion-cation contact along the cell edge , the ratio of the radius of the cation to that of anion is 0.731.

Thus,

\frac {r^+}{r^-}=0.731 .................1

Also, the sum of the radius of the cation and the anion in FCC is equal to half of the edge length.

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Given that:

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To find,

K^+\ (r^+) = ? \dot{A}

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4 years ago
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