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3 years ago

For the reaction H2(g) + 12(g) = 2HI(g), Kc = 50.2 at 445°C. If

Chemistry

1 answer:

Studentka2010 [4]3 years ago

5 0

Qc < Kc, the reaction proceeds from left to right to reach equilibrium

<h3>Further explanation </h3>

Given

K = 50.2 at 445°C

[H2] = [I2] = [HI] = 1.75 × 10⁻³ M At 445ºC

Reaction

H2(g) + I2(g) ⇔2HI(g)

Required

Solution

Qc for the reaction

$\tt Qc=\dfrac{[HI]^2}{[I_2][H_2]}\\\\Qc=\dfrac{(1.75.10^{-3})^2}{1.75.10^{-3}\times 1.75\.10^{-3}}=1$

Qc < Kc ⇒ reaction from left(reactants) to right (products) (the reaction will shift on the right) until it reaches equilibrium (Qc = Kc)

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A gas bottle contains 8. 61×1023 oxygen molecules at a temperature of 359. 0 k. what is the thermal energy of the gas? (you migh

Cloud [144]

A gas bottle contains 8. 61×1023 oxygen molecules at a temperature of 359. 0 k then the thermal energy of the gas is 6.35KJ.

<h3>What is thermal energy? </h3>

Thermal energy is the energy contained within a system which is responsible for its temperature. Heat is also termed as flow of thermal energy.

Thermal energy is directly proportional to the temperature. As temperature increase thermal energy increases.

Given,

temperature = 395K

Number of oxygen molecules= 8.61 × 10^(23)

Firstly we will calculate the number of moles = N/A

where A is the avagadro number = 6.022 × 10^(23)

number of moles = 8.61 × 10^(23)/6.022 ×10^(23)

number of moles = 1.42moles

Using the energy equation,

E = 3/2 × nRT

where R is the gas constant = 8.314J/molK

E = 3/2 × 1.42 × 8.314 × 359

= 6357.4J

= 6.35KJ

Thus, we found that the thermal energy of the gas is 6.35KJ.

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5 0

1 year ago

How many moles of HCl are present in 1.30 L of a 0.50 M solution

Shtirlitz [24]

Answer:

0.65moles of HCl was produced

4 0

2 years ago

determine the freezing point depression of a solution that contains 30.7 g glycerin (c3h8o3, molar mass

Blababa [14]

The freezing point depression of a solution containing 30.7 g of glycerin is calculated as -1.65°C

Equating :

It is given that,

Given mass of glycerin is = 30.7 grams (Solute)

Volume of water = 376 mL

$K_{f}$ or molar -freezing-depression point is = 1.86°C/m

Molar mass of glycerin = 92.09 g/mole

Now, to work out the value, the mass of water should be known. Thus, to calculate, the formula used will be:

Mass = Density X Volume

Mass = 1.0 g/mL X 376 mL

Mass = 376 g or 0.376 Kg

Using the formula of melting point depression, the equation becomes:

Δ $T_{f}$ = i × $K_{f}$ ×m

T⁰- $T_{s}$ = $i *K_{f} *\frac{mass of glycerin}{molar mass of glycerin * mass of water in kg}$

in which,

Δ $T_{f}$ = change in freezing point

Δ $T_{s}$ = freezing point of solution that has to be find

ΔT° = freezing point of water ()

Since, glycerin is a non-electrolyte, the Van't Hoff factor will be 1.

Substituting the values in the above equation:

0⁰C₋T $_{s}$ = 1 ×1.86°C/m × $\frac{30.7}{92.09g/mol * 0.376kg}$

$T_{s}$ = -1.65°C

Thus, the freezing point depression of a solution is -1.65°C

<h2 />

Freezing point depression

Freezing point depression is a colligative property observed in solutions that results from the introduction of solute molecules to a solvent. The freezing points of solutions are all less than that of the pure solvent and is directly proportional to the molality of the solute

Is melting point elevation or depression?

Boiling point elevation is that the raising of a solvent's boiling point due to the addition of a solute. Similarly, melting point depression is the lowering of a solvent's freezing point due to the addition of a solute. In fact, because the boiling point of a solvent increases, its melting point decreases

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8 0

1 year ago

What do alkali metals react bad with

solmaris [256]

Explanation:

what do you mean by "bad"

6 0

2 years ago

Read 2 more answers

A sample of gaseous neon atoms at atmospheric pressure and 0 °c contains 2.69 * 1022 atoms per liter. the atomic radius of neon

omeli [17]

Explanation

Radius of neon atom : 69 pm = $69\times 10^{-12} m$

Volume occupied by the one atom: $\frac{4}{3}\pi r^3$

$=\frac{4}{3}\times 3.14\times(69\times 10^{-12} m)^3=1.37\times 10^{-30} m^3$

given that $2.69\times 10^{22}$ atoms are present in 1L

1 L = 0.001 $m^3$

The volume occupied by the $2.69\times 10^{22}$ neon atoms

$2.69\times 10^{22}\times 1.37\times 10^{-30} m^3=3.68\times 10^{-8} m^3$

Fraction of volume occupied by the neon atom:

$=\frac{3.68\times 10^{-8} m^3}{0.001 m^3}=3.68\times 10^{-11} m^3$

$3.68\times 10^{-11} m^3$

The fraction of of volume occupied by the neon atom is very less than the 1 L which indicates the presence of large amount of empty space between the atoms of the gas.

3 0

3 years ago