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Dmitriy789 [7]
2 years ago
8

a chemist dissolved an 11.9-g sample of koh in 100.0 grams of water in a coffee cup calorimeter. when she did so, the water temp

erature increased by 26.0 ∘c. based on this, how much heat energy was required to dissolve the sample of koh? assume the specific heat of the solution is 4.184 j/g⋅°c.
Chemistry
1 answer:
Snowcat [4.5K]2 years ago
4 0

The heat of solution is -51.8 kJ/mol

<h3>What is the heat of solution?</h3>

We know that in a calorimeter, there is no loss or gain of energy. It is a good example of a closed system.

Number of moles of KOH =  11.9-g/56 g/mol = 0.21 moles

Temperature rise = 26.0 ∘c

Mass of the water = 100.0 grams

Heat capacity =  4.184 j/g⋅°c

Then;

ΔH = mcθ

ΔH = 100g * 4.184 j/g⋅°c * 26.0 ∘c = 10.88 kJ

Heat of solution = -(10.88 kJ/ 0.21 moles) = -51.8 kJ/mol

Learn more about heat of solution:brainly.com/question/24243878

#SPJ1

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If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate
andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0
3 years ago
How many liters of solvent are needed to make a 6.5M solution with 34 mol of solute?
Dominik [7]

Answer:

5.231 L.

Explanation:

  • Molarity is the no. of moles of solute per 1.0 L of the solution.

<em>M = (no. of moles of KCl)/(Volume of the solution (L))</em>

<em></em>

M = 6.5 M.

no. of moles of solute = 34.0 mol,

Volume of the solution = ??? L.

∴ (6.5 M) = (34.0 mol)/(Volume of the solution (L))

∴ (Volume of the solution (L) = (34.0 mol)/(6.5 M) = 5.231 L.

6 0
3 years ago
Two students were conducting an experiment. They have two blocks of different metals and wanted to identify them. They started b
erik [133]

Answer:

this dosen't make sense

5 0
3 years ago
Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79
BARSIC [14]

Answer :  The mole fraction and partial pressure of CH_4,CO_2 and He gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of CH_4 = 1.79 mole

Moles of CO_2 = 1.20 mole

Moles of He = 3.71 mole

Now we have to calculate the mole fraction of CH_4,CO_2 and He gases.

\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267

and,

\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179

and,

\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}

\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554

Thus, the mole fraction of CH_4,CO_2 and He gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of CH_4,CO_2 and He gases.

According to the Raoult's law,

p_i=X_i\times p_T

where,

p_i = partial pressure of gas

p_T = total pressure of gas  = 5.78 atm

X_i = mole fraction of gas

p_{CH_4}=X_{CH_4}\times p_T

p_{CH_4}=0.267\times 5.78atm=1.54atm

and,

p_{CO_2}=X_{CO_2}\times p_T

p_{CO_2}=0.179\times 5.78atm=1.03atm

and,

p_{He}=X_{He}\times p_T

p_{He}=0.554\times 5.78atm=3.20atm

Thus, the partial pressure of CH_4,CO_2 and He gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0
3 years ago
A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called
grandymaker [24]
Answer:
             <span>A geometric isomer with two alkyl groups on the same side of the carbon-carbon double bond is called <em>cis</em> Isomer.

Explanation:
                   Geometric isomerism takes place about the double bond in alkenes when the alkyl groups are either situated at the same side (<em>cis</em>) or are situated opposite (<em>trans</em>) to each other.

Example:
               <em>cis</em>-2-Butene (highlighted red)

               <em>trans</em>-2-Butene (highlighted blue)</span>

5 0
3 years ago
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