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Ksivusya [100]
2 years ago
5

The solubility of N2 in water at a particular temperature and at a N2 pressure of 1 atm is 6.8 × 10–4 mol L–1. Calculate the con

centration of dissolved N2 in water under normal atmospheric conditions where the partial pressure of N2 is 0.78 atm.
Chemistry
1 answer:
grin007 [14]2 years ago
7 0

Answer:

The correct answer is 5.30 * 10^-4 mol per L.

Explanation:

Based on Henry's law, in a solution solubility of the gas is directly proportional to the pressure, that is, C is directly proportional to P. Here P is the pressure and C is the concentration of the dissolved gases.  

Therefore, it can be written as,  

C2/C1 = P2/P1

Here, C1 is 6.8 * 10^-4 mol/L, P1 is 1 atm and P2 is 0.78 atm, then the value of C2 obtained by putting the values in the equation,  

C2/(6.8*10^-4) = 0.78/1

C2 = 0.78 * 6.8*10^-4

C2 = 5.30 * 10^-4 mol per L.  

Hence, the concentration of dissolved nitrogen at 0.78 atm is 5.30*10^-4 mol/L.  

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First you should know that there is seven oxygen atoms in one Mn2O7
So 
2.00 moles of Mn2O7 contain 14.00 moles of oxygen...
Then you multiply this no. with Avagadro no....
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Number of moles= no. of particles/avagadro's no..
14.00×6.02×10²³=84.28 atoms of oxygen... 
7 0
3 years ago
A balloon filled with helium gas occupies 2.50 L at 25°C and 1.00 atm. When released, it rises to an altitude where the temperat
stira [4]

Answer:

8,19 L

Explanation:

The combined gas law is a equation that can be used <em>when the initial and final conditions -</em>pressure,volume,amount of moles,temperature-, <em>of a gas change</em> during a process. It can be written:

\frac{P1V1}{n1T1}=\frac{P2V2}{n2T2}

If the amount of the gas remains constant, then n1=n2 and we have:

\frac{P1V1}{T1}=\frac{P2V2}{T2}

For the problem we have:

<em>Note: When working with gases is important to use </em><em><u>absolute temperature values (°K, °K=°C+273,15):</u></em>

P1=1 atm, V1=2,5L, T1=25+273,15=298,15°K

P2=0,3 atm, V2=?, T2=20+273,15=293,15°K

V2=\frac{P1V1T2}{T1P2}=\frac{1atm*2,5L*293,15K}{298,15K*0,3atm}=8,19L

The new volume of the balloon is 8,19 L.

5 0
2 years ago
My teacher gave this assignment for homework, and I don’t get it. A little help please?
SSSSS [86.1K]

County: cell

State: Tissue

Country: Organ

Continent: Organ system

World: organism

(smallest to largest)

3 0
2 years ago
200mL of 4.98M of sodium chloride solution is added to an additional 532 ml of water what is the final molarity?
ELEN [110]

Answer:

M₂ = 1.9 M

Explanation:

Given data;

Volume of sodium chloride = 200 mL

Molarity of sodium chloride = 4.98 M

Volume of water = 532 mL

Final Molarity = ?

Solution:

M₁V₁ = M₂V₂

M₂ =  M₁V₁ /V₂

M₂ = 4.98 M × 200 mL / 532 mL

M₂ = 996 mL. M /532 mL

M₂ = 1.9 M

4 0
3 years ago
What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.
OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

E^o_{cell}=(+0.34V)-(-0.76V)

E^o_{cell}=1.1V

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}

E_{cell}=1.14V

Therefore, the voltage of the voltaic cell is 1.14 V

5 0
2 years ago
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