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2 years ago

What is the concentration of each ion in a 0.525 mol/L solution of magnesium chloride? mg=? cl=?

Chemistry

1 answer:

Vladimir [108]2 years ago

3 0

Answer:

Mg⁺² = 0.525 mol/L

Cl⁻ = 1.05 mol/L

Explanation:

The formula for magnesium chloride is MgCl₂, and in a solution the following process takes place:

MgCl₂ → Mg⁺² + 2Cl⁻

By analyzing the stoichiometric coefficients, we can tell that the concentration of Mg⁺² ions is the same as the concentration of MgCl₂, while the Cl⁻ concentration is twice that value.

In other words:

Mg⁺² concentration = MgCl₂ concentration = 0.525 mol/L

Cl⁻concentration = 2 * MgCl₂ concentration = 1.05 mol/L

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Varvara68 [4.7K]

Answer:

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It is important to take into account that the pressure in the columns is a function of the height of the fluid (water or air).

Explanation:

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2 years ago

What is the oxidation number of phosphorus (P) in sodium phosphate (Na3PO4)?

joja [24]

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3 years ago

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A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

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2 years ago

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Paladinen [302]

Answer:

C6H12O6

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Explanation:

Glucose is a monocarbohydrate

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2 years ago

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2 years ago

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