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3 years ago

5

A chemist prepares a solution of silver nitrate by measuring out of silver nitrate into a volumetric flask and filling the flask

to the mark with water. calculate the concentration in of the chemist's silver nitrate solution. round your answer to significant digits.

1 answer:

Snezhnost [94]3 years ago

8 0

Amount of silver nitrate taken = 269.μmol $AgNO_{3}$

Volume of the solution = 300. mL

Concentration of a solution is generally expressed in terms of molarity. Molarity is defined as the moles of a substance present per liter of the solution.

$Molarity = \frac{Moles of solute}{Volume of solution(L)}$

We want the concentration in millimoles/L.

Converting μmol to millimol solute:

$269.$ μ $mol * \frac{1 millimol}{1000 micromol}$ = 0.269 millimol

Volume from mL to L: $300. mL * \frac{1 L}{1000 mL} = 0.300 L$

Therefore concentration of the chemist's solution = $\frac{0.269 millimol}{0.300 L} = 0.897 \frac{millimol}{L}$

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If I mix two chemicals in a test tube and notice after a few seconds that there are ice crystals forming on the outside of the t

miss Akunina [59]

Answer:

endothermic reaction

Explanation:

It simply means that you are witnessing<u> an endothermic reaction</u>.

An endothermic reaction is one that absorbs heat energy from its surrounding, thereby leaving the reaction vessel with a lower temperature as compared to before the reaction.

It is as opposed to exothermic reactions which are reactions that give off energy in the form of heat to the surrounding, thereby leaving a reaction vessel warmer than before the reaction.

<em>In this case, the formation of ice crystals outside the test tube means that heat energy has been absorbed by the reaction which leaves the vessel a temperature cold enough to activate the formation of ice. </em>

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3 years ago

The decomposition of dinitrogen pentoxide, N2O5, to NO2 and O2 is a first-order reaction. At 60°C, the rate constant is 2.8 × 10

Ugo [173]

Answer:

The correct option is a.

Explanation:

$2N_2O_5\rightarrow 4NO_2 + O_2$

125 kPa

125kpa - 2x 4x x

Total pressure after reaction = 176 kPa

125 kPa - 2x + 4x + x = 176 kPa

x = 17

125 kpa - 2x = 125 kPa - 2(17) = 91 kPa

Initial pressure of the dinitrogen pentoxide ,(at t=0) = $P_o= 125 kPa$

Final pressure of the dinitrogen pentoxide, (at t = t) = P = 91 kPa

The rate constant is = $k = 2.8\times 10^{-3} min^{-1}$

$t=\frac{2.303}{k}\log\frac{P_o}{P}$

$t=\frac{2.303}{2.8\times 10^{-3} min^{-1}}\log\frac{125 kPa}{91 kPa}$

$t=113.3969 minutes\approx 113 minutes$

It will take 113 minutes for the total pressure to reach 176 kPa.

8 0

3 years ago

If you mix 50mL of 0.1 M TRIS acid with 60 mL of0.2 M<br> TRIS base, what will be the resulting pH?

Katyanochek1 [597]

<u>Answer:</u> The pH of resulting solution is 8.7

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

<u>For TRIS acid:</u>

Molarity of TRIS acid solution = 0.1 M

Volume of solution = 50 mL

Putting values in above equation, we get:

$0.1M=\frac{\text{Moles of TRIS acid}\times 1000}{50mL}\\\\\text{Moles of TRIS acid}=0.005mol$

<u>For TRIS base:</u>

Molarity of TRIS base solution = 0.2 M

Volume of solution = 60 mL

Putting values in above equation, we get:

$0.2M=\frac{\text{Moles of TRIS base}\times 1000}{60mL}\\\\\text{Moles of TRIS base}=0.012mol$

Volume of solution = 50 + 60 = 110 mL = 0.11 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[salt]}{[acid]})$

$pH=pK_a+\log(\frac{[\text{TRIS base}]}{[\text{TRIS acid}]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of TRIS acid = 8.3

$[\text{TRIS acid}]=\frac{0.005}{0.11}$

$[\text{TRIS base}]=\frac{0.012}{0.11}$

pH = ?

Putting values in above equation, we get:

$pH=8.3+\log(\frac{0.012/0.11}{0.005/0.11})\\\\pH=8.7$

Hence, the pH of resulting solution is 8.7

6 0

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BlackZzzverrR [31]

Answer:

$n= \frac{m}{ \rho }* \frac{1 mol}{22.4 L}$

Explanation:

Assuming that all caculations are at normal pressure and -1.72°C :

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Where

$n$ is the number of moles of hydrogen

$n$ is the mass of hydrogen

$\rho$ is the density of hydrogen

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