Earth and the Solar System: Phenomena:

50cm3 of 1 mol/dm3 HCl at 30°C was mixed with 50cm3 of 1mol/dm3 NaOH at 30°C in a styrofoam calorimeter. The temperature of the

trapecia [35]

Answer:

-21 kJ·mol⁻¹

Explanation:

Data:

H₃O⁺ + OH⁻ ⟶ 2H₂O

V/mL: 50 50

c/mol·dm⁻³: 1.0 1.0

ΔT = 4.5 °C

C = 4.184 J·°C⁻¹g⁻¹

C_cal = 50 J·°C⁻¹

Calculations:

(a) Moles of acid

$\text{Moles of acid} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}\\\\\text{Moles of base} = \text{0.050 dm}^{3} \times \dfrac{\text{1.0 mol}}{\text{1 dm}^{3}} = \text{0.050 mol}$

So, we have 0.050 mol of reaction

(b) Volume of solution

V = 50 dm³ + 50 dm³ = 100 dm³

(c) Mass of solution

$\text{Mass of solution} = \text{100 dm}^{3} \times \dfrac{\text{1.00 g}}{\text{1 dm}^{3}} = \text{100 g}$

(d) Calorimetry

There are three energy flows in this reaction.

q₁ = heat from reaction

q₂ = heat to warm the water

q₃ = heat to warm the calorimeter

q₁ + q₂ + q₃ = 0

nΔH + mCΔT + C_calΔT = 0

0.050ΔH + 100×4.184×4.5 + 50×4.5 = 0

0.050ΔH + 1883 + 225 = 0

0.050ΔH + 2108 = 0

0.050ΔH = -2108

ΔH = -2108/0.0500

= -42 000 J/mol

= -42 kJ/mol

This is the heat of reaction for the formation of 2 mol of water

The heat of reaction for the formation of mol of water is -21 kJ·mol⁻¹.

5 0

3 years ago

When substances are formed that are different from the starting substances, a physical change has occurred. True False

Sergio039 [100]

False, because physical change does not involve the formation of substances. An example of physical change is change of state e.g water freezing .It is still water if you were to melt it.

4 0

3 years ago

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What is the mass of flourine-19

Arisa [49]

18.998403

Explanation:
The internet told me

7 0

3 years ago

A slice of pizza has 500 kcal. If we could burn the pizza and use all the heat to warm a 50-L container of cold water, what woul

gayaneshka [121]

Answer:

10°C