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3 years ago

20. Which of the following is a method of purifying water? A. Adding iodine B. Freezing C. Aeration D. Fixation

Chemistry

2 answers:

Oxana [17]3 years ago

7 0

The answer is A, adding Iodine

Dmitry_Shevchenko [17]3 years ago

6 0

If you add 8-10 drops of iodine in water it will purify it.

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The voltage generated by the zinc concentration cell described by the line notation zn(s) ∣∣ zn2+(aq,0.100 m) ∥∥ zn2+(aq,? m) ∣∣

lubasha [3.4K]

Reaction involved in present electrochemical cell,
At Anode: Zn → Zn^2+ + 2e^2-
At cathode: Zn^2+ + 2e^2- → Zn
Net Reaction: Zn + Zn^2+ ('x' m) → Zn^2+(0.1 m) + Zn
Number of electrons involved in present electrochemical cell = n = 2

According to Nernst equation for electrochemical cell,
Ecell = -2.303 $\frac{RT}{nT} log \frac{[Zn^2+]R}{[Zn^2+]L}$ = 0.014

Given: T = $25^{0}C = 298 K$ , F = 96500 C, R = gas constant = 8.314J/K.mol, [Zn^2+]R = 0.1 m , Ecell = 0.014 v

∴ 0.014 = - 2.303 $\frac{8.314X298}{2X96500}log \frac{0.1}{x}$
∴ log $\frac{0.1}{x}$ = $\frac{-2.303X8.314X298}{2X96500X0.014}$ = -2.1117
∴ log x = log(0.1) + 2.1117
∴x = 13.09 m

3 0

3 years ago

Read 2 more answers

What is the Lewis dot diagram for GaF3 Can someone help me

s344n2d4d5 [400]

Answer:

View attached image

8 0

3 years ago

What kind of data would you need to collect to carry out this experiment?

JulijaS [17]

You need to specify the experiment to get an answer. However, most experiments require qualitative data (descriptive, non-numerical data) and quantitative data (measured, numerical data).

4 0

4 years ago

Calcium has a cubic closest packed structure as a solid. Assuming that calcium has an atomic radius of 197 pm, calculate the den

NISA [10]

Answer:

$\rho=1.54\ g/cm^3$

Explanation:

The expression for density is:

$\rho=\frac {Z\times M}{N_a\times {{(Edge\ length)}^3}}$

$N_a=6.023\times 10^{23}\ {mol}^{-1}$

M is molar mass of Calcium = 40.078 g/mol

For cubic closest packed structure , Z= 4

$\rho$ is the density

Radius = 197 pm = $1.97\times 10^{-8}\ cm$

Also, for fcc, $Edge\ length=2\sqrt{2}\times radius=2\sqrt{2}\times 1.97\times 10^{-8}\ cm=5.572\times 10^{-8}\ cm$

Thus,

$\rho=\frac{4\times \:40.078}{6.023\times \:10^{23}\times \left(5.572\times 10^{-8}\right)^3}\ g/cm^3$

$\rho=\frac{160.312}{10^{23}\times \:6.023\left(10^{-8}\times \:5.572\right)^3}\ g/cm^3$

$\rho=\frac{160.312}{10^{23}\times \:1.04195E-21}\ g/cm^3$

$\rho=\frac{160.312}{104.19483}\ g/cm^3$

$\rho=1.54\ g/cm^3$

7 0

3 years ago

The boiling point of an aqueous solution is 101.02 °C. What is the freezing point?

kompoz [17]

Colligative properties calculations are used for this type of problem. Calculations are as follows:

ΔT(boiling point) = 101.02 °C - 100.0 °C= 1.02 °C
ΔT(boiling point) = (Kb)m
m = 1.02 °C / 0.512 °C kg / mol
m = 1.99 mol / kg

ΔT(freezing point) = (Kf)m
ΔT(freezing point) = 1.86 °C kg / mol (1.99 mol / kg)
ΔT(freezing point) = 3.70 °C
Tf - T = 3.70 °C
T = -3.70 °C

3 0

4 years ago