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3 years ago

How many moles of magnesium chloride (mgcl2) are in 1.31 l of 0.38 m mgcl2?

1 answer:

5 0

Molarity can be defined as the number of moles dissolved in 1 L of solution.
The molarity of MgCl₂ is 0.38 mol/L
In 1 L of MgCl₂ solution, amount go 0.38 moles is dissolved.
Therefore in 1.31 L of MgCl₂ - 0.38 mol /1 L x 1.31 L = 0.4978 mol rounded off is 0.50 mol
In 1.31 L of solution, amount of 0.50 mol of MgCl₂ is present.

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3. Given 20g of Barium Hydroxide, how many grams of

anastassius [24]

The number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

First, we will write the balanced chemical equation for the reaction

The balanced chemical equation for the reaction is

Ba(OH)₂ + 2NH₄NO₃ → 2NH₄OH + Ba(NO₃)₂

This means, 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

Now, we will calculate the number of moles of barium hydroxide present.

Mass of barium hydroxide (Ba(OH)₂) = 20 g

Using the formula

$Number\ of\ moles = \frac{Mass}{Molar\ mass}$

Molar mass of Ba(OH)₂ = 171.34 g/mol

∴ Number of moles of Ba(OH)₂ present = $\frac{20}{171.34}$

Number of moles of Ba(OH)₂ present = 0.116727 mole

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Since 1 mole of barium hydroxide is required to react with 2 moles of ammonium nitrate

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0.116727 mole of barium hydroxide will react with 2 × 0.116727 mole of ammonium nitrate

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∴ Number of moles of NH₄NO₃ required is 0.233454 mole

Now, for the mass of ammonium nitrate (NH₄NO₃) required

From the formula

Mass = Number of moles × Molar mass

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∴ Mass of NH₄NO₃ required = 0.233454 × 80.043

Mass of NH₄NO₃ required = 18.68636 g

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Hence, the number of grams of ammonium nitrate (NH₄NO₃) it would take for all the barium hydroxide to react is 18.7g

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Answer:

Explanation:

The complete question is shown in the image attached.

Let us call to mind the fact that the SN1 mechanism involves the formation of carbocation in the rate determining step. The order of stability of cabocations is; tertiary > secondary > primary > methyl.

Hence, a tertiary alkyl halide is more likely to undergo nucleophilic substitution reaction by SN1 mechanism since it forms a more stable cabocation in the rate determining step.

Structure IV is a tertiary alkyl halide, hence it is more likely to undergo nucleophilic substitution reaction by SN1 mechanism.

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