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melamori03 [73]
3 years ago
6

What is a nanomaterial that is often used with other compounds to desalinate and decontaminate water?

Chemistry
1 answer:
Katarina [22]3 years ago
3 0

Answer:

Silver ions.

Explanation:

The most extensively studied nanomaterials for water purification and treatment mainly includes zero-valent metal nanoparticles, metal oxides nanoparticles, carbon nanotubules and nanocomposites.

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Hydrogen reacts with iodine to form hydrogen iodide: H2[g] + 12[g] = 2H1[g] The
ELEN [110]

Answer:

follow my instagram to help you with your school work in general

6 0
3 years ago
A buret was improperly read to 1 decimal place giving a reading of 27.2 mL. If the actual volume is 27.26 mL, what is the percen
Elenna [48]
Using the significant figure it would be 27.3
8 0
3 years ago
4NH3+5O2-4NO+6H2O<br>how many moles of NH3 must react to produce 5.0 moles of NO?​
tresset_1 [31]

Answer: 5.0 moles

Explanation:

From the equation, we see that for every 4 moles of ammonia consumed, 4 moles of nitrogen monoxide are produced (we can reduce this to moles of ammonia consumed = moles of nitrogen monoxide produced).

This means that the answer is <u>5.0 mol</u>

7 0
2 years ago
For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coeffici
Brut [27]

<u>Question:</u>

For the cell constructed from the hydrogen electrode and metal-insoluble salt electrode, B) calculate the mean activity coefficient for 0.124 b HCl solution if E=0.342 V at 298 K

<u>Answer:</u>

The mean activity coefficient for HCl solution is 0.78.

<u>Explanation:</u>

Activity coefficient is defined as the ratio of any chemical activity of any substance with its molar concentration. So in an electrochemical cell, we can write activity coefficient as γ. The equation for determining the mean activity coefficient is

              E=E_{0}-0.0514 \mathrm{V} \ln \gamma

As we know that E_{0} = 0.22 V and E = 0.342 V, the equation will become

             0.342 V+0.0514 V \ln (0.124)=0.22 V-0.0514 V \ln \gamma

             0.342 V-0.222 V=-0.0514 V(\ln \gamma+\ln 0.124)

             0.12 \mathrm{V}=-0.0514 \mathrm{V}(\ln \gamma+\ln 0.124)

             \frac{0.12}{0.0514}=-\ln (0.124 \gamma)

             -2.3346=\ln (0.124 \gamma)

             e^{-2.3346}=0.124 \gamma

             0.0968=0.124 \gamma

             \gamma=\frac{0.0968}{0.124}=0.78

So, the mean activity coefficient is 0.78.

6 0
3 years ago
Find the density of a material given that a 5.03g sample occupies 3.24ml
Verizon [17]

Formula for density:

D = Mass / Volume

Mass = 5.03g

Volume = 3.24ml


Formula in our case:

D = 5.03 / 3.24

D = 0.31~g/m^{3}


Hope it helped,


BioTeacher101

3 0
4 years ago
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