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DaniilM [7]
2 years ago
11

A scientist produces zinc iodide (zni2). (a) calculate the minimum mass of zinc that needs to be added to 0.500 g of iodine so t

hat the iodine fully reacts. the equation for the reaction is: zn + i2 ⟶ zni2 relative atomic masses (mr): zn = 65 i = 127
Chemistry
1 answer:
harina [27]2 years ago
4 0

The minimum mass of Zinc, Zn that needs to be added to 0.500 g of iodine so that the iodine fully reacts is 0.128 g

<h3>Balanced equation </h3>

Zn + I₂ —> ZnI₂

Molar mass of Zn = 65 g/mol

Mass of Zn from the balanced equation = 1 × 65 = 65 g

Molar mass of I₂ = 127 × 2 = 254 g/mol

Mass of I₂ from the balanced equation = 1 254 = 254 g

SUMMARY

From the balanced equation above,

254 g of I₂ required 65 g of Zn

<h3>How to determine the mass of Zn needed </h3>

From the balanced equation above,

254 g of I₂ required 65 g of Zn

Therefore,

0.5 g of I₂ will require = (0.5 × 65) / 254 = 0.128 g of Zn

Thus, the minimum mass of Zn required is 0.128 g

Learn more about stoichiometry:

brainly.com/question/14735801

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Free_Kalibri [48]

Nh4cl is sometimes preferred instead of hcl or h2so4 for "acid" work-up after grignard reactions, particularly when the expected and desired product is a tertiary alcohol because NH4+ is a much milder acid than HCl or H2SO4, which achieve the protonation of

the oxyanion to yield the alcohol while minimizing the risk of dehydration.

Ammonium chloride (NH4Cl) is the used as reagent that quenches the magnesium alkoxide product of the Grignard addition.

It is a proton source without being acidic as in acidic medium the protonation of the tertiary alcohol product and elimination to the alkene.

In the presence of HCl or any other strong acid protonation proceed and form alkene but not with ammonium chloride.

Thus from above we concluded that Nh4cl is preferred instead of hcl or h2so4 for "acid" work-up after grignard reactions.

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6 0
1 year ago
A 100.0 mL sample of 0.10 M Ca(OH)2 is titrated with 0.10 M HBr. Determine the pH of the solution after the addition of 300.0 mL
Vera_Pavlovna [14]

Answer : The correct option is, (C) 1.7

Explanation :

First we have to calculate the moles of Ca(OH)_2 and HBr.

\text{Moles of }Ca(OH)_2=\text{Concentration of }Ca(OH)_2\times \text{Volume of solution}=0.10M\times 0.1L=0.01mole

\text{Moles of }HBr=\text{Concentration of }HBr\times \text{Volume of solution}=0.10M\times 0.3L=0.03mole

The balanced chemical reaction will be:

Ca(OH)_2+2HBr\rightleftharpoons CaBr_2+2H_2O

0.01 mole of Ca(OH)_2 dissociate to give 0.01 mole of Ca^{2+} ion and 0.02 mole of OH^- ion

and

0.03 mole of HBr dissociate to give 0.03 mole of H^+ ion and 0.03 mole of Br^- ion

That means,

0.02 moles of OH^- ion  neutralize by 0.02 moles of H^+ ion.

The excess moles of H^+ ion = 0.03 - 0.02 = 0.01 mole

Total volume of solution = 100 + 300 = 400 ml = 0.4 L

Now we have to calculate the concentration of H^+ ion.

\text{Concentration of }H^+=\frac{\text{Moles of }H^+}{\text{Total volume}}

\text{Concentration of }H^+=\frac{0.01mole}{0.4L}=0.025M

Now we have to calculate the pH of the solution.

pH=-\log [H^+]

pH=-\log (0.025M)

pH=1.7

Therefore, the pH of the solution is, 1.7

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