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gayaneshka [121]
2 years ago
11

Calculate the ratio of effusion rates of cl2 to f2 .

Chemistry
1 answer:
Lelechka [254]2 years ago
4 0
<span>Answer: Graham's law of gaseous effusion states that the rate of effusion goes by the inverse root of the gas' molar mass. râšM = constant Therefore for two gases the ratio rates is given by: r1 / r2 = âš(M2 / M1) For Cl2 and F2: r(Cl2) / r(F2) = âš{(37.9968)/(70.906)} = 0.732 (to 3.s.f.)</span>
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A particular radioactive isotope has a half-life of (2.50+a) hours. if you have (24.5+b) g of the isotope at 10:00 am, how much
gavmur [86]
Its a not b and i dont know the reast

3 0
2 years ago
The reaction SO2(g)+2H2S(g)←→3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2 from power-plant stack gases. T
kifflom [539]

Answer : The equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

Explanation :

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

First we have to calculate the standard free energy of reaction (\Delta G^o).

\Delta G^o=G_f_{product}-G_f_{reactant}

\Delta G^o=[n_{S(s)}\times \Delta G_f^0_{(S(s))}+n_{H_2O(g)}\times \Delta G_f^0_{(H_2O(g))}]-[n_{SO_2(g)}\times \Delta G_f^0_{(SO_2(g))}+n_{H_2S(g)}\times \Delta G_f^0_{(H_2S(g))}]

where,

\Delta G^o = standard free energy of reaction = ?

n = number of moles

Now put all the given values in this expression, we get:

\Delta G^o=[3mole\times (0kJ/mol)+2mole\times (-228.57kJ/mol)]-[1mole\times (-300.4kJ/mol)+2mole\times (-33.01kJ/mol)]

\Delta G^o=-90.72kJ/mol

Now we have to calculate the value of K_p

\Delta G^o=-RT\ln K_p

where,

\Delta G_^o =  standard Gibbs free energy  = -90.72 kJ/mol

R = gas constant = 8.314 J/mole.K

T = temperature = 298 K

K_p = equilibrium constant  = ?

Now put all the given values in this expression, we get:

-90.72kJ/mol=-(8.314J/mol.K)\times (298K) \ln K_p

K_p=7.98\times 10^{15}

Now we have to calculate the value of K_p.

The given balanced chemical reaction is,

SO_2(g)+2H_2S(g)\rightleftharpoons 3S(s)+2H_2O(g)

The expression for equilibrium constant will be :

K_p=\frac{(p_{H_2O})^2}{(p_{H_2S})^2\times (p_{SO_2})}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Let the equilibrium SO_2 pressure be, x

Pressure of SO_2 = Pressure of H_2S = x

Now put all the given values in this expression, we get

7.98\times 10^{15}=\frac{(22)^2}{(x)^2\times (x)}

x=3.93\times 10^{-5}torr

Thus, the equilibrium SO_2 pressure is, 3.93\times 10^{-5}torr

4 0
2 years ago
A gas is heated from 263.0 K to 298.0 K and the volume is increased from 24.0 liters to 35.0 liters by moving a large piston wit
Triss [41]

Answer:

The final pressure is approximately 0.78 atm

Explanation:

The original temperature of the gas, T₁ = 263.0 K

The final temperature of the gas, T₂ = 298.0 K

The original volume of the gas, V₁ = 24.0 liters

The final volume of the gas, V₂ = 35.0 liters

The original pressure of the gas, P₁ = 1.00 atm

Let P₂ represent the final pressure, we get;

\dfrac{P_1 \cdot V_1}{T_1} = \dfrac{P_2 \cdot V_2}{T_2}

P_2 = \dfrac{P_1 \cdot V_1 \cdot T_2}{T_1 \cdot V_2}

P_2 = \dfrac{1 \times 24.0 \times 298}{263.0 \times  35.0} = 0.776969038566

∴ The final pressure P₂ ≈ 0.78 atm.

4 0
2 years ago
A mixture of carbon dioxide and an unknown gas was allowed to effuse from a container. The carbon dioxide took 1.25 times as lon
Aleks04 [339]

Answer:

CO

Explanation:

From Graham's law, time taken to diffuse is directly proportional to the molecular mass of the gases. For two different gases.

t1/t2=√m1/m2

Since gas 1 diffuse 1.25 times as slowly as gas 2 and gas 1 is CO2 with m as 44g

1.25/1=√44/m2

Therefore m2=28g CO

7 0
2 years ago
The molar mass of oxygen gas (O2) is 32.00 g/mol. The molar mass of C3H8 is 44.1 g/mol. What mass of O2, in grams, is required t
Ira Lisetskai [31]
The reaction formula of this is C3H8 + 5O2 --> 3CO2 + 4H2O. The ratio of mole number of C3H8 and O2 is 1:5. 0.025g equals to 0.025/44.1=0.00057 mole. So the mass of O2 is 0.00057*5*32=0.0912 g.
6 0
2 years ago
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