This question is based on Dalton's Law of Partial Pressure which states that "the total pressure of a system of gas is equal to the sum of the pressure of each individual gas (partial pressure).
Now, Partial Pressure of a gas = (mole fraction) × (total pressure)
⇒ Partial Pressure of Hydrogen =
×
= 0.31 atm
Thus the Partial Pressure of Hydrogen in the container is
0.31 atm.
Answer:
Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M
Explanation:
4A + 3B ------> C + 2D
In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s
The amount of A that has reacted at the end of 3 seconds will be
0.08 × 3 = 0.24 M
Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.
From the chemical reaction,
4 moles of A gives 1 mole of C
0.24 M of reacted A will form (0.24 × 1)/4 M of C
Amount of C formed at the end of the 3s interval = 0.06 M
If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.
If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M
Answer: The results agree with the law of conservation of mass
Explanation:
The law of conservation of mass states that mass is neither created nor destroyed in a chemical reaction. On the reactant side, the total mass of reactants is 14.3g and the total product masses is also 14.3g. That implies that no mass was !most in the reaction. The sum of masses on the left hand side corresponds with sum of masses on the right hand side of the reaction equation.
Answer:
823.7g
Explanation:
Using the formula as follows:
Q = m × c × ∆T
Where;
Q = amount of heat (J)
m = mass of substance (g)
c = specific heat capacity (J/g°C)
∆T = change in temperature (°C)
Using the information given in this question as follows:
Q = 6,400 J
m = ?
c of soil = 0.840 J/g°C
∆T = 9.25°C
Using Q = mc∆T
m = Q ÷ c∆T
m = 6,400 ÷ (0.840 × 9.25)
m = 6400 ÷ 7.77
m = 823.7g
