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Ilia_Sergeevich [38]
3 years ago
8

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<br><br> 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
<span>ΔT(boiling point)  = (Kb)m
</span>m =  1.02 °C / 0.512 °C kg / mol
<span>m = 1.99 mol / kg

</span><span>ΔT(freezing point)  = (Kf)m
</span>ΔT(freezing point)  = 1.86 °C kg / mol (<span>1.99 mol / kg)
</span>ΔT(freezing point)  = 3.70 <span>°C
</span>Tf - T = 3.70 <span>°C
T = -3.70 </span><span>°C</span>
3 0
4 years ago
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