Answer
- continuous removal of PH3
- adding more of P into the system
Explanation:
In the reaction P4(g)+6H2(g) ⇌ 4PH3(g);
- The effect of temperature on equilibrium has to do with the heat of reaction. Recall that for an endothermic reaction, heat is absorbed in the reaction, and the value of ΔH is positive. Thus, for an endothermic reaction, we can picture heat as being a reactant:
heat+A⇌BΔH=+
- Since the reaction is endothermic reaction, heat is a absorbed. Decreasing the temperature will shift the equilibrium to the left, while increasing the temperature will shift the equilibrium to the right forming more of PH3.
- According to Le Chatelier’s principle, adding additional reactant to a system will shift the equilibrium to the right, towards the side of the products. In the same Way, reducing the concentration of the product will also shift equilibrium to the right continually forming PH3 as it is removed.
Answer:
The mass of xenon in the compound is 2.950 grams
Explanation:
Step 1: Data given
Mass of XeF4 = 4.658 grams
Molar mass of XeF4 = 207.28 g/mol
Step 2: Calculate moles of XeF4
Moles XeF4 = mass XeF4 / molar mass XeF4
Moles XeF4 = 4.658 grams / 207.28 g/mol
Moles XeF4 = 0.02247 moles
Step 3: Calculate moles of xenon
XeF4 → Xe + 4F-
For 1 mol xenon tetrafluoride, we have 1 mol of xenon
For 0.02247 moles XeF4 we have 0.02247 moles Xe
Step 4: Calculate mass of xenon
Mass xenon = moles xenon * molar mass xenon
Mass xenon = 0.02247 moles * 131.29 g/mol
Mass xenon = 2.950 grams
The mass of xenon in the compound is 2.950 grams
The answer is oxygen. (02)
Answer:
V¹N²= V²N²
here V¹= ?
N¹= 6.00
V²= 175ml
M²= 0.2M
So V¹= (V²N²)/N² = (175 x 0.2)/6
V¹ = 5.83 ml
Explanation:
Therefore diluting 5.83 ml of 6.00M NaOH to 175 m l ,we get 0.2M Solution.
Answer:
The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.
Explanation:
Mass of compound X = 7.00 g
Moles of compound X = 
Mass of water in calorimeter ,m= 35.00 kg = 35000 g
Change in temperature of the water in calorimeter = ΔT
ΔT = 2.113°C
Specific heat capacity of water ,c= 4.186 J/g °C
Q = m × c × ΔT

Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.
Heat of formation of Compound X at 25°C:

= -3095.75 kJ/mol