A solution made from pure Potassium hydroxide contained 1.85 g of Potassium hydroxide in exactly 200 cm3 of water. Using phenolphthalein indicator, titration of 20.0 cm3 of this solution is carried out v/s sulphuric acid. 9.35 cm3 of sulphuric acid solution is required for complete neutralisation. [atomic masses: K= 39, O = 16, H = 1]
(a) write the equation for the titration reaction.
(b) calculate the molarity of the Potassium hydroxide solution. (c) calculate the moles of Potassium hydroxide neutralised. (d) calculate the moles of sulphuric acid neutralised.
(e) calculate the molarity of the sulphuric acid.

Answers

Answer 1

The concentration of H2SO4 is  0.176 M. The number of moles of H2SO4 is 0.0165 moles.

The equation of the reaction is;

2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + H2O(l)

The number of moles of KOH = 1.85 g/56 g/mol = 0.033 moles

concentration = number of moles /volume

volume = 200 cm3 or 0.2 L

concentration of KOH = 0.033 moles/0.2 L = 0.165 M

From the titration formula;

CAVA/CBVB =na/nb

CA = concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

na = number of moles of acid

nb = number of moles of base

Making CA the subject of formula;

CA = CBVBna/VAnb

CA = ?

CB = 0.165 M

VA =  9.35 cm3

VB = 20.0 cm3

na = 1

nb =2

CA = CBVBna/VAnb

CA = 0.165 M ×  20.0 cm3  × 1/9.35 cm3 × 2

CA = 0.176 M

The concentration of sulphuric acid is 0.176 M

From the reaction equation;

2 moles of KOH reacts with 1 mole of H2SO4

0.033 moles of KOH reacts with 0.033 moles ×  1 mole/2 moles

= 0.0165 moles of H2SO4

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6

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What is radioactive decay?

Unstable radioactive isotopes of atoms undergo nuclear decay by the emission of charged particles such alpha or beta particles. Radioactive decay is a first order reaction. Thus, the decay constant is written as:

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Answers

Answer:

About 10.2 g Li₃N.

Explanation:

We are given the reaction:

[tex]\displaystyle \text{6 Li(s) + N$_2$(g) }\longrightarrow \text{2 Li$_3$N(s)}[/tex]

And we want to determine the amount of Li₃N that can be formed from 6.07 g of Li and an excess of nitrogen.

To convert from g Li to g Li₃N, we can: (1) convert from g Li to mol Li, (2) mol Li to mol Li₃N, and (3) mol Li₃N to g Li₃N.

The molecular weight of Li is 6.94 g/mol. From the equation, six moles of Li yields two moles of Li₃N. And the molecular weight of Li₃N is 34.83 g/mol as shown below.

Molecular weight of Li₃N:

[tex]\displaystyle \begin{aligned}\text{MW}_\text{Li$_3$N} & = (3 (6.94) + 14.01) \text{ g/mol} \\ \\ & =34.83\text{ g/mol} \end{aligned}[/tex]

This yields three ratios:

[tex]\displaystyle \frac{1 \text{ mol Li}}{6.94 \text{ g Li}}, \frac{2 \text{ mol Li$_3$N}}{6 \text{ mol Li}}, \text{ and } \frac{34.83 \text{ g Li$_3$N}}{1 \text{ mol Li$_3$N}}[/tex]

From the initial value, multiply:

[tex]\displaystyle 6.07 \text{ g Li} \cdot \frac{ 1 \text{ mol Li}}{6.94 \text{ g Li}} \cdot \frac{2 \text{ mol Li$_3$N}}{6 \text{ mol Li}} \cdot \frac{34.83 \text{ g Li$_3$N}}{1 \text{ mol Li$_3$N}} = 10.2\text{ g Li$_3$N}[/tex]

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Hello there!

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Rock cycle(Attachment-1)

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Density is the amount of space that atoms are compacted in a substance.Apr 25, 2017

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Answers

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