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laila [671]
4 years ago
11

1 kg of water (specific heat = 4184 J/(kg K)) is heated from freezing (0°C) to boiling (100°C). What is the change in thermal en

ergy?
Chemistry
1 answer:
Andrews [41]4 years ago
5 0

Answer: 1560632 joules

Explanation:

The change in thermal energy (Q) required to heat ice depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)

Thus, Q = MCΦ

Given that:

Q = ?

Mass of frozen water (ice) = 1kg

C = 4184 J/(kg K)

Φ = (Final temperature - Initial temperature)

= 100°C - 0°C = 100°C

Convert 100°C to Kelvin

(100°C + 273) = 373K

Then, Q = MCΦ

Q = 1kg x 4184 J/(kg K) x 373K

Q = 1560632 joules

Thus, the change in thermal energy is 1560632 joules

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Answer: 1. 9.08\times 10^{-6} moles

2. 90 mg

Explanation:

O_3(g)+2NaI(aq)+H_2O(l) \rightarrow O_2(g)+I_2(s)+2NaOH(aq)

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 4.54 \times 10^{-6} moles of ozone is removed by =\frac{2}{1}\times 4.54 \times 10^{-6}=9.08\times 10^{-6} moles of sodium iodide.

Thus 9.08\times 10^{-6} moles of sodium iodide are needed to remove 4.54\times 10^{-6} moles of O_3

2. \text{Number of moles of ozone}=\frac{0.01331g}{48g/mol}=0.0003moles

According to stoichiometry:

1 mole of ozone is removed by 2 moles of sodium iodide.

Thus 0.0003 moles of ozone is removed by =\frac{2}{1}\times 0.0003=0.0006 moles of sodium iodide.

Mass of sodium iodide= moles\times {\text {molar mass}}=0.0006\times 150g/mol=0.09g=90mg    (1g=1000mg)

Thus 90 mg of sodium iodide are needed to remove 13.31 mg of O_3.

3 0
3 years ago
What is the most important use of an elements atomic number
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Answer:

Atoms of each element contain a characteristic number of protons and electrons. The number of protons determines an element's atomic number and is used to distinguish one element from another.

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Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha
nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L

The moles of air are:

3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}

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