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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
Chemistry
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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2.1. The lithium ion in a 250,00 mL sample of mineral water was
juin [17]

Answer:

11482 ppt of Li

Explanation:

The lithium is extracted by precipitation with B(C₆H₄)₄. That means moles of Lithium = Moles B(C₆H₄)₄. Now, 1 mole of B(C₆H₄)₄ produce the liberation of 4 moles of EDTA. The reaction of EDTA with Mg²⁺ is 1:1. Thus, mass of lithium ion is:

<em>Moles Mg²⁺:</em>

0.02964L * (0.05581mol / L) = 0.00165 moles Mg²⁺ = moles EDTA

<em>Moles B(C₆H₄)₄ = Moles Lithium:</em>

0.00165 moles EDTA * (1mol B(C₆H₄)₄ / 4mol EDTA) = 4.1355x10⁻⁴ mol B(C₆H₄)₄ = Moles Lithium

That means mass of lithium is (Molar mass Li=6.941g/mol):

4.1355x10⁻⁴ moles Lithium * (6.941g/mol) = 0.00287g. In μg:

0.00287g * (1000000μg / g) = 2870μg of Li

As ppt is μg of solute / Liter of solution, ppt of the solution is:

2870μg of Li / 0.250L =

<h3>11482 ppt of Li</h3>

4 0
2 years ago
How many atoms are in 80.45 g of magnesium?
KatRina [158]

Hello!

To find the amount of atoms that are in 80.45 grams of magnesium, we will need to know Avogadro's number and the mass of one mole of magnesium.

Avogadro's number is 6.02 x 10^23 atoms, and one mole of magnesium is equal to 24.31 grams.

1. Divide by one mole of magnesium

80.45 / 24.31 = 3.309 moles (rounded to the number of sigfigs)

2. Multiply moles by Avogadro's number

3.309 x (6.02 x 10^23) = 1.99 x 10^24 (rounded to the number of sigfigs)

Therefore, there are 1.99 x 10^24 atoms in 80.45 grams of magnesium.

8 0
2 years ago
You can identify a metal by carefully determining its density. A 8.44 g piece of an unknown metal is 1.25 cm long, 2.50 cm wide,
Dafna1 [17]

Answer: Aluminum, 2.70 g/cm^3

Explanation:

Density is defined as the mass contained per unit volume.  It is characteristic of a substance.

Density=\frac{mass}{Volume}

Given : Mass of object = 8.44 grams

Volume of object=length\times breadth\times height=1.25cm\times 2.50cm\times 1.0cm=3.125cm^3

Putting in the values we get:

Density=\frac{8.44g}{3.125cm^3}=2.70g/cm^3

Thus density of the object will be 2.70 g/cm^3  which matches that of aluminium.

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Answer:

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