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ANEK [815]
3 years ago
11

n a coffee cup calorimeter, 100.0 mL of 1.0M NaOH and 100.0 mL of 1.0M HCl are mixed. Both solutions were originally at 24.6 deg

rees Celsius. After the reaction, the final temperature is 31.3 degrees Celsius. Assuming that all the solution have a density of 1.0g/cm3 and a specific heat capacity of 4.18 J/oCg, calculate the enthalpy change for the neutralization of HCl by NaOH. Assume that no heat is lost to the surroundings or to the calorimeter.
Chemistry
1 answer:
RoseWind [281]3 years ago
3 0

Answer:

Explanation:

mass of the solution = volume x density = 200 x 1 = 200 gm

heat absorbed = m x s x Δ t , s is specific heat , Δt is rise in temperature

= 200 x 4.18 x ( 31.3 - 24.6 )

= 5601 J .

This is the enthalpy change required.

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2 years ago
Suppose of potassium sulfate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of potassium
dimulka [17.4K]

Answer:

This question is incomplete, here's the complete question:

<em><u>"Suppose 0.0842g of potassium sulfate is dissolved in 50.mL of a 52.0mM aqueous solution of sodium chromate. Calculate the final molarity of potassium cation in the solution. You can assume the volume of the solution doesn't change when the potassium sulfate is dissolved in it. Round your answer to 2 significant digits."</u></em>

Explanation:

Reaction :-

K2SO4 + Na2CrO4 ------> K2CrO4 + Na2SO4

Mass of K2SO4 = 0.0842 g, Molar mass of K2SO4 = 174.26 g/mol

Number of moles of K2SO4 = 0.0842 g / 174.26 g/mol = 0.000483 mol

Concentration of Na2CrO4 = 52.0 mM = 52.0 * 10^-3 M = 0.052 mol/L

Volume of Na2CrO4 solution = 50.0 ml = 50 L / 1000 = 0.05 L

Number of moles of Na2CrO4 = 0.05 L * 0.052 mol/L = 0.0026 mol

Since number of moles of K2SO4 is smaller than number of moles Na2CrO4, so 0.000483 mol of K2SO4 will react with 0.000483 mol of Na2CrO4 will produce 0.000483 mol of K2CrO4.

0.000483 mol of K2CrO4 will dissociate into 2* 0.000483 mol of K^+

Final concentration of potassium cation

= (2*0.000483 mol) / 0.05 L = 0.02 mol/L = 0.02 M

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Answer:

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Explanation:

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