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2 years ago

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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

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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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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The depth of the lake is 67.164 meters.

Explanation:

The combined gas equation is,

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

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$P_1$ = 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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