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STALIN [3.7K]
4 years ago
7

¿Cómo moléculas de un sólido difieren de las de un líquido o un gas?

Chemistry
1 answer:
castortr0y [4]4 years ago
8 0
Ssssssssssssssssssssssssssss
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A 101.2 ml sample of 1.00 m naoh is mixed with 50.6 ml of 1.00 m h2so4 in a large styrofoam coffee cup; the cup is fitted with a
Murrr4er [49]

The enthalpy change of the reaction when sodium hydroxide and sulfuric acid react can be calculated using the mass of solution, temperature change, and specific heat of water.

The balanced chemical equation for the reaction can be represented as,

H_{2}SO_{4}(aq) + 2NaOH (aq) ----> Na_{2}SO_{4}(aq) + 2H_{2}O(l)

Given volume of the solution = 101.2 mL + 50.6 mL = 151.8 mL

Heat of the reaction, q = m C .ΔT

m is mass of the solution = 151.8 mL * \frac{1 g}{1 mL} = 151.8 g

C is the specific heat of solution = 4.18 \frac{J}{g. ^{0}C}

ΔT is the temperature change = 31.50^{0}C - 21.45^{0}C = 10.05^{0}C

q = 151.8 g (4.18 \frac{J}{g ^{0}C})(10.05^{0}C) = 6377 J

Moles of NaOH = 101.2 mL * \frac{1L}{1000 mL}*\frac{1.00 mol}{L} = 0.1012 mol NaOH

Moles of H_{2}SO_{4} = 50.6 mL * \frac{1 L}{1000 mL} * \frac{1.0 mol}{1 L} = 0.0506 mol H_{2}SO_{4}

Enthalpy of the reaction = \frac{6377 J*\frac{1kJ}{1000J}}{0.0506 mol} = 126 kJ/mol

5 0
3 years ago
What is it called when a substance changes from a vapor to a liquid
skad [1K]
The answer is CONDENSATION.
4 0
3 years ago
The amount of heat absorbed when 5.5g of aluminum is heated from 25.0c to 95.0c the specific heat of aluminum is 0.897J/(gc)​
NemiM [27]

Answer:

Q=21.99 kJ

Explanation:

4 0
3 years ago
The diagram below represents the rock cycle.
spayn [35]

Answer:

Heat and Pressure

Explanation:

4 0
3 years ago
When 3.93 grams of lactic acid, CHoOs(s), are burned in a bomb
aliya0001 [1]

The heat released in the combustion of lactic acid is absorbed by the

calorimeter and in the decomposition of the lactic acid.

ΔH°f of lactic acid is approximately <u>-716.2 kJ</u>

Reasons:

Known parameters are;

Mass of the lactic acid = 3.93 grams

Heat  capacity of the bomb calorimeter = 10.80 kJ·K⁻¹

Change in temperature of the calorimeter, ΔT = 5.34 K

ΔHrxn = ΔErxn

ΔH°f of H₂O(l) = -285.8 kJ·mol⁻¹

ΔH°f of CO₂(g) = -393.5 kJ·mol⁻¹

The chemical equation for the reaction is presented as follows;

  • C₃H₆O₃ + 2O₂ → 3CO₂ + 3H₂O

The heat of the reaction = 10.80 kJ·K⁻¹ × 5.34 K = 57.672 kJ

Molar mass of C₃H₆O₃ = 90.07 g/mol

Number of moles of C₃H₆O₃ = \dfrac{3.93 \, g}{90.07 \, g/mol} = 0.043633 moles

Number of moles of CO₂ produced = 3 × 0.043633 moles = 0.130899 moles

Heat produced = 0.130899 mole × -285.8 kJ·mol⁻¹ = -37.4109342 kJ

Moles of H₂O produced = 0.130899 moles

Heat produced = 0.130899 mole × -393.5 ≈ -51.51 kJ

Therefore, we have;

Heat absorbed by the lactic acid = ΔH°f of H₂O + ΔH°f of CO₂ + Heat absorbed by the calorimeter

Which gives;

Heat absorbed by lactic acid  = -37.4109342 kJ - 51.51 kJ + 57.672 kJ ≈ -31.249 kJ

The heat absorbed by the lactic acid ≈ -31.249 kJ

  • \Delta H^{\circ}f \ of \ C_3H_6O_3 = \dfrac{-31.249}{0.043633} \approx  -716.2

ΔH°f of C₃H₆O₃ ≈ -716.2 kJ

Heat of formation of lactic acid ≈ <u>-716.2 kJ</u>.

Learn more here:

brainly.com/question/13185938

5 0
3 years ago
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