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3 years ago

Determine the concentration of a solution prepared by diluting 20.0 mL of 2.00 M NaCl to 250.0 mL. Please show your work.

Chemistry

1 answer:

yawa3891 [41]3 years ago

6 0

0.16 M is the concentration of a solution prepared by diluting 20.0 ml of 2.00 M NaCl to 250.0 ml.

Explanation:

Data given:

Initial volume of NaCl, V1 = 20 ml

initial molarity of the NaCl solution = 2M

Final volume of the NaCl solution = 250 ml

final molarity of the diluted solution = ?

from the information given, the formula for dilution used is:

Minitial Vinitial = Mfinal Vfinal

putting the values in the rearranged equation:

V final = $\frac{Minitial Vinitial }{Vfinal}$

V final = $\frac{20 X 2}{250}$

Mfinal = 0.16 M

Thus it can be seen that when a 20 ml solution having molarity of 2M is diluted to 250 ml the molarity decreases to 0.16 M.

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ss7ja [257]

Answer:

The steps are explained below, the essential step is to find mass here, 120 g of NaOH.

Explanation:

In order to answer this question, we need to define molarity conceptually firstly to see what variables we need. According to the formula, molarity is equal to the ratio between moles and volume, while moles itself is a ratio between mass and molar mass. This means we have a formula for molarity involving mass, molar mass and volume:

$c = \frac{n}{V} = \frac{m}{MV}$

In order to prepare a 500.0 mL of stock solution of 6.0 M of NaOH, we then need to find the mass of NaOH dissolved in this solution using the equation above:

$m = cMV = 6.0 M\cdot 39.997 g/mol\cdot 0.5000 L = 120 g$

Now, since we have the mass of NaOH, we can describe the steps needed to prepare this solution:

measure 120 grams of solid NaOH;
add this mass of NaOH into a 500.0-mL Erlenmeyer flask;
fill approximately half of the flask with distilled water and stir gently to make sure that NaOH dissolves, if it doesn't, add more water and repeat the process;
when NaOH fully dissolves, fill the flask to the mark.

Our solution is prepared.

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4 years ago

The standard free energy of activation of a reaction A is 81.9 kJ mol–1 (19.6 kcal mol–1) at 298 K. Reaction B is one million ti

klasskru [66]

Answer:

54.9 kJ/mol

Explanation:

The relation between the activation energy (Ea) and the rate constant (k) is given by the Arrhenius equation.

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where,

A is a collision factor

R is the ideal gas constant

T is the absolute temperature

Reaction B is one million times faster than reaction A at the same temperature. So $k_{B}=10^{6} k_{A}$ .

Then,

$k_{B}=10^{6} k_{A}\\A.e^{-Ea_{B}/RT}=10^{6}A.e^{-Ea_{A}/RT}\\e^{-Ea_{B}/RT}=10^{6}e^{-Ea_{A}/RT}\\ln(e^{-Ea_{B}/RT})=ln(10^{6}e^{-Ea_{A}/RT})\\\frac{-Ea_{B}}{RT} =ln10^{6} -\frac{Ea_{A}}{RT} \\Ea_{B}=(ln10^{6} -\frac{Ea_{A}}{RT}).(-RT)=(ln10^{6}-\frac{89.1kJ/mol}{(8.314\times 10^{-3} kJ/mol.K).298K} ).(-8.314\times 10^{-3} \frac{kJ}{mol.K}.298K )=54.9kJ/mol$

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The correct answer is A. The image shows a nuclear fission. This takes place in any of the heavy nuclei after capture of a neutron. This is the opposite of nuclear fusion. In this case, nuclei are broken down into two.

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nasty-shy [4]

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Let's consider the following reaction.

N₂(g) + O₂(g) → 2 NO(g)

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