Answer : The new pressure acting on a 2.5 L balloon is, 8.6 atm.
Explanation :
Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.
or,
where,
= initial pressure = 3.7 atm
= final pressure = ?
= initial volume = 5.8 L
= final volume = 2.5 L
Now put all the given values in the above equation, we get:
Thus, the new pressure acting on a 2.5 L balloon is, 8.6 atm.
Complete question:
Consider the reaction.
At equilibrium at 600 K, the concentrations are as follows.
2HF -----> H₂ + F₂
[HF] = 5.82 x 10-2 M
[H2] = 8.4 x 10-3 M
[F2] = 8.4 x 10-3 M
What is the value of Keq for the reaction expressed in scientific notation?
2.1 x 10-2
2.1 x 102
1.2 x 103
1.2 x 10-3
Answer:
2.1 × 10^-2
Explanation:
Kequilibrum(Keq) = product/reactant
Equation for the reaction :
2HF -----> H₂ + F₂
Therefore,
Keq = [H2][F2] / [HF]^2
Keq = [8.4 x 10-3][8.4 x 10-3] / [5.82 x 10-2]^2
Keq = [70.56 × 10^(-3 + - 3)]/[33.8724 × 10^(-2×2)]
Keq = [70.56 × 10^-6] / [33.8724 × 10^-4]
Keq = 2.0665 × 10^(-6 - (-4))
Keq = 2.0665 × 10^(-6 + 4)
Keq = 2.1 × 10^-2
99.6 mg= .0996 g
.0996g x (1mol/125.05g) x (1 mol SO3/1 mol Na2SO2) x (6.02 x 10^22/1mol SO3) = 4.79 x 10^19 SO32- ions
In this question we need to find the new volume of the gas. Since we have been given the pressure and temperature change, we can used to combined gas law equation.
the parameters for 1st instance are given on the left side and parameters for the second instance are given on the right side of the equation
(319 mmHg x 0.558 L)/ 115 K = (215 mmHg x V)/387 K
V = 2.79 L
Answer : The enthalpy change during the reaction is -6.48 kJ/mole
Explanation :
First we have to calculate the heat gained by the reaction.
where,
q = heat gained = ?
m = mass of water = 100 g
c = specific heat = 
= final temperature = 
= initial temperature = 
Now put all the given values in the above formula, we get:
Now we have to calculate the enthalpy change during the reaction.
where,
= enthalpy change = ?
q = heat gained = 23.4 kJ
n = number of moles barium chloride = 
Therefore, the enthalpy change during the reaction is -6.48 kJ/mole
