The volume of a sample of ammonia gas : 5.152 L
<h3>Further explanation</h3>
Given
0.23 moles of ammonia
Required
The volume of a sample
Solution
Assumed on STP
Conditions at T 0 ° C and P 1 atm are stated by STP (Standard Temperature and Pressure). At STP, Vm is 22.4 liters / mol.
So for 0.23 moles :
= 0.23 x 22.4 L
= 5.152 L
Answer:
Enthalpy of formation = -947.68KJ/mol
Explanation:
Enthalpy of formation is the heat change when one mole of a substance is formed from its element in its standard states and in standard conditions of temperature and pressure. it may be positive or negative, if positive, it is an endothermic reaction where the heat content of the product is greater than that of the reactants, and if negative, it is exothermic reaction - where the heat content of the reactants is greater than the products. the enthalpy of formation is measured in KiloJoule/Moles (KJ/Mole).
From the value of the enthalpy of formation of NaHCO3, it shows that the reaction is exothermic, that is the formation of NaHCO3 from its constituents elements. As such, the heat content of the reactants is greater than the products.
The step by step explanation is shown in the attachment.
Given that the volume and amount of water are kept constant,
P/T = constant
P₁/T₁ = P₂/T₂
Normal atmospheric pressure is 746 mmHg and normal boiling point of water is 100 °C.
746/100 = 589/T₂
T₂ = 79.0 °C
Answer:
<h2>Ammonia Gas</h2>
Explanation:
It result in formation of ammonia gas.
N2 + 3H2 ---<u>iron</u><u>/</u><u>molybdenum</u><u>/</u><u>high</u><u> </u>temp/pres--- > 2 NH3
It forms ammonia gas.
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Answer : The value of
at this temperature is 66.7
Explanation : Given,
Pressure of
at equilibrium = 0.348 atm
Pressure of
at equilibrium = 0.441 atm
Pressure of
at equilibrium = 10.24 atm
The balanced equilibrium reaction is,

The expression of equilibrium constant
for the reaction will be:

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


Therefore, the value of
at this temperature is 66.7