What is the boiling point elevation and the boiling point of a 1.50 m solution of NaCl

What is the percent yield of a reaction in which 200. g of phosphorus trichloride reacts with excess water to form 91.0 g of hcl

Snowcat [4.5K]

Answer is: <span>yield of a reaction is 56,4%.
</span>Chemical reaction: PCl₃ + 3H₂O → 3HCl + H₃PO₃.
m(PCl₃) = 200 g.
m(HCl) = 91,0 g.
n(PCl₃) = m(PCl₃) ÷ M(PCl₃).
n(PCl₃) = 200 g ÷ 137,33 g/mol.
n(PCl₃) = 1,46 mol.
n(HCl) = m(HCl) ÷ M(HCl).
n(HCl) = 91 g ÷ 36,45 g/mol.
n(HCl) = 2,47 mol.
From reaction: n(PCl₃) : n(HCl) = 1 : 3.
n(HCl) = 1,46 mol · 3 = 4,38 mol.
Yield of reaction: 2,47 mol ÷ 4,38 mol · 100% = 56,4%.

4 0

3 years ago

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A chemist wants to combine sodium with a halogen. The halogen must have a higher atomic mass than sodium. Which of the following

user100 [1]

Answer:

bromine

Explanation:

i just know

8 0

3 years ago

Exactly how many -OH groups are found in a 6 carbon alditol?

mr Goodwill [35]

Answer:

The number of OH group found in found in a 6 carbon alditol is 6

Explanation:

The structure of a 6 carbon alditol is shown on the first uploaded image

8 0

3 years ago

The theoretical yield of a reaction is the amount of product obtained if the limiting reactant is completely converted to produc

Elis [28]

Answer: 9.9 grams

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

a) moles of $H_2$

$\text{Number of moles}=\frac{8.150g}{2g/mol}=4.08moles$

b) moles of $C_2H_4$

$\text{Number of moles}=\frac{9.330g}{28g/mol}=0.33moles$

$H_2(g)+C_2H_4(g)\rightarrow C_2H_6(g)$

According to stoichiometry :

1 mole of $C_2H_4$ combine with 1 mole of $H_2$

Thus 0.33 mole of $C_2H_4$ will combine with = $\frac{1}{1}\times 0.33=0.33$ mole of $H_2$

Thus $C_2H_4$ is the limiting reagent as it limits the formation of product.

As 1 mole of $C_2H_4$ give = 1 mole of $C_2H_6$

Thus 0.33 moles of $C_2H_4$ give = $\frac{1}{1}\times 0.33=0.33moles$ of $C_2H_6$

Mass of $C_2H_6=moles\times {\text {Molar mass}}=0.33moles\times 30g/mol=9.9g$

Thus theoretical yield (g) of $C_2H_6$ produced by the reaction is 9.9 grams

7 0

3 years ago

When a strong acid is titrated with a strong base using phenolphthalein, the color changes suddenly at the end point. The color

larisa86 [58]

Answer:

Keto - enol tautomerism

Explanation:

Phenolphthalein is an organic molecule (formula: C20H14O4), which is used in to determine the final volume in an acid-base titration, this is, as the molecule changes from one to another color, depending on the Ph of the solution (in acids solution, the molecule remains colorless, while in basics solutions it remains pink) :

The molecule of phenolphthalein is a week acid, which losses a proton (H+) when it´s in solution: the undissociated molecule is colorless, while the correspondent anion (without a H+) is pink. This may be simplified with the following reaction:

H₃In⁺ ⇄ H₂In (colorless) ⇄ In⁻² (pink) ⇄ In(OH) ⁻³

- When adding a base (for example, NaOH) to phenolphthalein (initially, a colorless solution), molecule losses one H⁺ and we obtain In⁻², which is pink

- When adding an acid, the original structured is recovered, so color turns from pink to colorless

This change color is not only explained with adding an acid or a base, but also with phenolphthalein structure, that leads to a keto-enol tautomerism: as molecule has 2 hydroxyl groups (- OH) and 1 carbonyl group (C = O), compounds with this structure have an equilibrium between both groups:

R1 = C H– OH ⇄ R1 – C = O

With this change, phenolphthalein structure changes and hence, color solution changes as well

4 0

2 years ago