We can write the balanced equation for the synthesis reaction as
H2(g) + Cl2(g) → 2HCl(g)
We use the molar masses of hydrogen chloride gas HCl and hydrogen gas H2 to calculate for the mass of hydrogen gas H2 needed:
mass of H2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol H2 / 2 mol HCl) *
(2.02 g H2 / 1 mol H2)
= 4.056 g H2
We also use the molar masses of hydrogen chloride gas HCl and chlorine gas CL2 to calculate for the mass of hydrogen gas H2:
mass of CL2 = 146.4 g HCl *(1 mol HCl / 36.46 g HCl) * (1 mol Cl2 / 2 mol HCl) *
(70.91 g Cl2 / 1 mol Cl2)
= 142.4 g Cl2
Therefore, we need 4.056 grams of hydrogen gas and 142.4 grams of chlorine gas to produce 146.4 grams of hydrogen chloride gas.
Explanation:
Given
The enthalpy of formation of RbF (s) is –557.7kJ/mol
The standard enthalpy of formation of RbF (aq, 1 m) is –583.8 kJ/mol
The enthalpy of solution of RbF = Enthalpy of RbF (aq) - Enthalpy of formation of RbF (s)
= -583.8 - (-557.7) kJ/mol
= -26.1 kJ/mol
The enthalpy is negative which means that the temperature will rise when RbF is dissolved.
<span>Van der waal or ideal eqn is given by PV = NRT; P = NRT/ V.
Where N = 1.335 is the number of moles. T = 272K is temperature. V = 4.920L is the volume. And R = 0.08205L. Substiting the values into the eqn; we have,
P = (1.331* 0.08205 * 272)/ 4.920 = 29.7047/ 4.920 = 6.03atm.</span>
Answer:
P₂ = 3.86 atm .
Explanation:
We assume that during this change , the temperature of the gas remains constant .
So the gas will obey Boyle's law .
P₁ V₁ = P₂V₂
5.08 x 7.56 = P₂ x 9.94
P₂ = 3.86 atm .
Answer:
571.81 mL
Explanation:
Assuming constant pressure, we can solve this problem by using <em>Charles' law</em>, which states that at constant pressure:
Where in this case:
We <u>input the data</u>:
- 852 mL * 200 K = V₂ * 298 K
And <u>solve for V₂</u>:
The new volume would be 571.81 mL.