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3 years ago

After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2 is

Chemistry

1 answer:

Tanzania [10]3 years ago

4 0

Answer:

Ksp = 8.8x10⁻⁵

Explanation:

<em>Full question is:</em>

<em>After mixing an excess PbCl2 with a fixed amount of water, it is found that the equilibrium concentration of Pb2+ is 2.8 × 10–2 M. What is Ksp for PbCl2?</em>

<em />

When an excess of PbCl₂ is added to water, Pb²⁺ and Cl⁻ ions are produced following Ksp equilibrium:

PbCl₂(s) ⇄ Pb²⁺ + 2Cl⁻

Ksp = [Pb²⁺] [Cl⁻]²

If an excess of PbCl₂ was added, an amount of Pb²⁺ is produced (X) and twice Pb²⁺ is produced as Cl⁻ (2X):

Ksp = [X] [2X]²

Ksp = 4X³

As X is the amount of Pb²⁺ = 2.8x10⁻²M:

Ksp = 4(2.8x10⁻²)³

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I need help with this, please :00000

mojhsa [17]

theoretical yield of the reaction is 121.38 g of NH₃ (ammonia)

limiting reactant is N₂ (nitrogen)

excess reactant is H₂ (hydrogen)

Explanation:

We have the following chemical reaction:

N₂ + 3 H₂ → 2 NH₃

Now we calculate the number of moles of each reactant:

number of moles = mass / molar weight

number of moles of N₂ = 100 / 28 = 3.57 moles

number of moles of H₂ = 100 / 2 = 50 moles

From the chemical reaction we see that 3 moles of H₂ are reacting with 1 moles of N₂, so 50 moles of H₂ are reacting with 16.66 moles of N₂ but we only have 3.57 moles of N₂ available, so the limiting reactant will be N₂ and the excess reactant will be H₂.

Knowing the chemical reaction and the limiting reactant we devise the following reasoning:

if 1 mole of N₂ produce 2 moles of NH₃

then 3.57 moles of N₂ produce X moles of NH₃

X = (3.57 × 2) / 1 = 7.14 moles of NH₃

mass = number of moles × molar weight

mass of NH₃ = 7.14 × 17 = 121.38 g

theoretical yield of the reaction is 121.38 g of NH₃

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limiting reactant

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4 0

3 years ago

Show all calculations. 1. 2 C 4 H 10 + 13 O 2 -> 8 CO 2 + 10 H 2 O a) what mass of O 2 will react with 400 g C 4 H 10? b) how

bazaltina [42]

$\\ \tt\hookrightarrow 2C_4H_10+13O_2\longrightarrow 8CO_2+10H_2O$

13mol of O_2 reacts with 2mols of C_4H_10.
7.5mol of O_2 reacts with 1mol of C_4H_10

No of moles:-

$\\ \tt\hookrightarrow \dfrac{Given\:mass}{Molar\:mass}$

$\\ \tt\hookrightarrow \dfrac{400}{58}$

$\\ \tt\hookrightarrow 6.89mol$

Now

Moles of O_2

$\\ \tt\hookrightarrow 7.5(6.89)=51.6mol$

Mass of O_2

$\\ \tt\hookrightarrow 51.6(32)=1651.2g$

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2 years ago

Which label belongs in the region marked X?

lubasha [3.4K]

Answer:

A) involves changes in temperature

Explanation:

The figure is missing, but I assume that the region marked X represents the region in common between Gay-Lussac's law and Charle's Law.

Gay-Lussac's law states that:

"For an ideal gas kept at constant volume, the pressure of the gas is directly proportional to its absolute temperature"

Mathematically, it can be written as

$p\propto T$

where p is the pressure of the gas and T its absolute temperature.

Charle's Law states that:

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Mathematically, it can be written as

$V\propto T$

where V is the volume of the gas and T its absolute temperature.

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