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-Dominant- [34]

2 years 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

1 answer:

Nesterboy [21]2 years 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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Given these reactions, where X represents a generic metal or metalloid 1) H2(g)+12O2(g)⟶H2O(g)ΔH1=−241.8 kJ 1) H2(g)+12O2(g)⟶H2O

Bond [772]

Answer:

ΔH = -793,6 kJ

Explanation:

It is possible to obtain ΔH of this reaction using Hess's law that says you can sum the half-reactions ΔH to obtain the ΔH of the global reaction:

If half-reactions are:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The sum of (4) + 4×(3) - (2) - 2×(1) - 2×(5) is:

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369,2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356,9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483,6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= <em>XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)</em>

Where ΔH is:

ΔH = -639,1 kJ -369,2 kJ -356,9 kJ +483,6 kJ +88,0 kJ

<em>ΔH = -793,6 kJ</em>

I hope it helps!

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2 years ago

Which of the following statements concerning a metal crystallized in a face-centered cubic cell is/are CORRECT? 1. One metal ato

AleksAgata [21]

Answer:

Only 3 is correct.

Explanation:

The crystal of a metal or an ionic compound is called a cell, and there are 7 types of unit cells: cubic, tetragonal, orthorhombic, monoclinic, hexagonal, rhombohedral, and triclinic.

In a face-centered cubic cell (FCC) all angles are 90º and all lengths are equal. Each cubic cell has 8 atoms in each corner of the cube, and that atom is shared with 8 neighboring cells. So for a metal crystal, the atom is located at each of the eight lattice points, where it is shared equally between eight unit cells.

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Use the conversion factor 1amu=6.66054 x 10^-24 to answer the following questions.. . a) 1.674 x 10^-24 g of neutrons is how man

Wewaii [24]

A.) The conversion factor is 1 amu = $1.66054 ^{-24}g$

To know how many amu in $1.674x10^{-24}g$ grams of neutrons:
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=1.00811 amu

b.) The mass in grams of one lithium ion which has an atomic weight of 6.94 amu.

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c.) How many amu in 6.492x10^-23g potassium?

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2 years ago

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A chemist is working on a reaction represented by this chemical equation:

erastova [34]

Answer is: A. 1.81 mol.

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n(FeCl₂) = 4.15 mol; amount of iron(II) chloride.

n(KOH) = 3.62 mol; amount of potassium hydroxide, limiting reactant.

From chemical reaction: n(KOH) : n(Fe(OH)₂) = 2 : 1.

n(Fe(OH)₂) = n(KOH) ÷ 2.

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2 years ago

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igor_vitrenko [27]

Answer:

1 M

Explanation:

Magnesium chloride will furnish chloride ions as:

$MgCl_2\rightarrow Mg^{2+}+2Cl^-$

Given :

Moles of magnesium chloride = 0.20 mol

Thus, moles of chlorine furnished by magnesium chloride is twice the moles of magnesium chloride as shown below:

$Moles =2\times 0.20\ moles$

Moles of chloride ions by magnesium chloride = 0.40 moles

Potassium chloride will furnish chloride ions as:

$KCl\rightarrow K^{+}+Cl^-$

Given :

Moles of potassium chloride = 0.10 moles

Thus, moles of chlorine furnished by potassium chloride is same as the moles of potassium chloride as shown below:

Moles of chloride ions by potassium chloride = 0.10 moles

Total moles = 0.40 + 0.10 moles = 0.50 moles

Given, Volume = 500 mL = 0.5 L (1 mL = 10⁻³ L)

Concentration of chloride ions is:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

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2 years ago