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ad-work [718]
2 years ago
14

A 50.0 mL sample of 0.10 M potassium nitrate is mixed with 100 mL of 0.25 M magnesium nitrate. What is the final concentration o

f nitrate in the solution
Chemistry
1 answer:
ser-zykov [4K]2 years ago
7 0

The final concentration of nitrate is 0.367 M

<h3>Concentration of a substance</h3>

The concentration of a substance is the amount of substance present in a given volume of a solution of that substance.

Concentration = number of moles/ volume

<h3>Equation for the dissociation of potassium nitrate</h3>

KNO3 ---> K+ + NO3-

1 mole of KNO3 produces 1 mole of nitrate

moles of nitrate in a 0.1 M KNO3 solution

  • moles = concentration × volume

volume of KNO3 = 50.0 mL= 0.05 L

moles of nitrate = 0.1 × 0.05

moles of nitrate in KNO3 = 0.005

<h3>Equation for the dissociation magnesium nitrate</h3>
  • Mg(NO3)2 ---> Mg^2+ + 2 NO3-

1 mole of Mg(NO3)2 produces 2 moles of nitrate ion

moles of Mg(NO3)2 = concentration × volume

volume of solution = 100 mL= 0.1 L

moles of Mg(NO3)2 = 0.1 × 0.25 = 0.025

Moles of nitrate = 0.025 × = 0.05

Total moles of nitrate = 0.05 + 0.005 = 0.055 moles

Total volume of solution = 0.1 + 0.05 = 0.15 L

Concentration of nitrate = 0.055/0.15

Concentration of nitrate = 0.367 M

Therefore, the final concentration of nitrate is 0.367 M

Learn more about concentration and solutions at: brainly.com/question/17206790

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Answer: Hydrogen molecules will have greatest average speed.

Explanation:

The formula for average speed is  :

\mu_{av}={\sqrt{\frac{8RT}{\pi\times M}}}

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Now putting all the values:

\frac{\nu_{H_2}}{\nu_{O_2}}=\sqrt{\frac{8RT}{\pi\times M_{H_2}}}/{\sqrt{\frac{8RT}{\pi\times M_{O_2}}}

\frac{\nu_{H_2}}{\nu_{O_2}}=\sqrt{\frac{16}{2}}

\frac{\nu_{av}_{H_2}}{\nu_{av}_{O_2}}=8

Thus average speed of hydrogen is 8 times the average speed of oxygen. Thus hydrogen molecules will have greatest average speed

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The volume of a bubble that starts at the bottom of a lake at 4.55°C increases by a factor of 8.00 as it rises to the surface wh
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Answer:

The depth of the lake is 67.164 meters.

Explanation:

The combined gas equation is,

\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}

where,

P_1 = initial pressure of gas in bubble= ?

P_2 = final pressure of gas = 0.980 atm

V_1 = initial volume of gas = V

V_2 = final volume of gas = 8.00 × V

T_1 = initial temperature of gas = 4.55^oC=273.15+4.55=277.7 K

T_2 = final temperature of gas = 18.05^oC=273.15+18.05=291.2 K

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

\frac{P_1\times V}{277.7 K}=\frac{0.980 atm\times 8.00\times V}{291.2 K}

P_1=7.476 atm

pressure of the gas in bubble initially is equal to the sum of final pressure and pressure exerted by water at depth h.

P_1=P_2+h\rho\times g

Where :

\rho = density of water = 1.00 g /cm^3=1000 g/m^3

g = acceleration due gravity = 9.8 m/s^2

7.476 atm=0.980 atm +h\rho\times g

6.496 atm=h\rho\times g

6.496 \times 101325 Pa=h\1000 g/m^3\times 9.8 m/s^2

h = 67.164 m

The depth of the lake is 67.164 meters.

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3 years ago
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