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-Dominant- [34]
1 year ago
12

2. Suggest four ways in which the concentration of PH3 could be increased in an equilibrium described by the following equation:

P4(g)+6H2(g) ⇌ 4PH3(g) ΔH=110.5kJ
Chemistry
1 answer:
Nesterboy [21]1 year ago
4 0

Answer

  • increase in temperature
  • decrease in pressure
  • 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.

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What is the mass of solute in 200.0 L of a 1.556-M solution of KBr
sergejj [24]
The molarity of KBr solution is 1.556 M
molarity is defined as the number of moles of solute in volume of 1 L solution.
the number of KBr moles in 1 L - 1.556 mol
Therefore in 200.0 L - 1.556 mol/L x 200.0 L = 311.2 mol
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4 0
1 year ago
If 42.1 mL of 1.02 M sodium hydroxide, measured using a graduated cylinder, is placed in a beaker filled with 300 mL of DI water
lys-0071 [83]

Answer:

Approximately 0.126 M

Explanation:

For the calculation of the dilution you take into account the moles of NaOH in the 42.1mL of the original solution and you use the new volume of 342.1 mL:

(42.1 mL * 1.02 M ) Number Of Moles\\\\C=\frac{42.1mL*1.02M}{342.1 mL} = 0.126 M

The standardization is necessary because a beaker is not not an instrument used to measure volumes and the marks on it only give an estimate of the volume of the solution, they are used to contain solutions and carry reactions among other things. If you would have measured the water with a graduated cylinder (an instrument designed to measure volumes) the standardization wouldnt be that necessary.

5 0
2 years ago
In a titration experiment, H2O2(aq) reacts with aqueous MnO4-(aq) as represented by the equation above. The dark purple KMnO4 so
pickupchik [31]

Answer:

C. 4x10⁻⁴ mol / (Ls)

Explanation:

Based in the reaction:

5 H₂O₂(aq) + 2 MnO₄⁻(aq) + 6 H⁺(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 O₂(g)

2 moles of MnO₄⁻ disappears while 5 moles of O₂ appears.

If 5 moles appears in a rate of 1.0x10⁻³mol /(Ls), 2 moles will disappear:

2 moles ₓ (1.0x10⁻³mol /(Ls) / 5 moles) = <em>4x10⁻⁴ mol / (Ls)</em>

Right answer is:

C. 4x10⁻⁴ mol / (Ls)

8 0
2 years ago
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