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3241004551 [841]
3 years ago
5

How long is 5 half lives if the half life is 2.6 hours? Remember to round to the correct number of significant figures and use u

nits ( hours)
Chemistry
1 answer:
Dmitry [639]3 years ago
3 0

Answer:

\Delta t = 13\,h

Explanation:

Half time is period required to desintegrating the half of the initial number of atoms. Then, the total time is:

\Delta t = 5\cdot (2.6\,h)

\Delta t = 13\,h

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2. Nitric oxide contains 46.66% nitrogen and 53.34% oxygen. Water contains 11.21% hydrogen and 88.79% oxygen. Ammonia contains 1
Mrrafil [7]

Answer:

The law of reciprocal proportions states that if two elements react individually with a given weight of a third element, the ratio of the masses with which they combine with the third element are either the same or a simple multiple of the ratio of the masses with which they combine with each other

The compounds formed includes;

1) Nitric oxide, NO

Nitrogen = 46.66% × 30.01 = 14

Oxygen = 53.34% × 30.01 = 16

2) Water, H₂O

Hydrogen = 11.21% × 18.01528 = 2

Oxygen = 88.79% × 18.01528 ≈ 16

3) Ammonia, NH₃

Hydrogen = 17.78% × 17.031 ≈ 3

Nitrogen = 82.22% × 17.031 ≈ 14

The ratio of nitrogen to oxygen in nitric oxide = 14:16 = 7:8

The ratio of nitrogen to hydrogen in ammonia = 14:3

The ratio in which hydrogen and oxygen combine with nitrogen = 3/16

The ratio of hydrogen and oxygen combine with each other in water = 2/16

Therefore, the ratio with which hydrogen and oxygen combine with nitrogen, is (2/3) times the ratio with which they combine with each other, which verifies the law of reciprocal proportions

Explanation:

4 0
3 years ago
When a 1.00-g sample of methane gas was burned with excess oxygen in the calorimeter, the temperature increased by 7.3°C. When
Advocard [28]

Answer:

The energies of  combustion (per gram) for hydrogen and methane are as follows: Methane = 82.5 kJ/g;  Hydrogen = 162 kJ/g

<em>Note: The question is incomplete. The complete question is given below:</em>

To compare the energies of combustion of these fuels, the  following experiment was carried out using a bomb  calorimeter with a heat capacity of 11.3 kJ/℃.  When a 1.00-g sample of methane gas burned with

<em>excess oxygen in the calorimeter, the temperature  increased by 7.3℃. When a 1.00 g sample of  hydrogen gas was burned with excess oxygen, the temperature increase was 14.3°C. Compare the energies of  combustion (per gram) for hydrogen and methane.</em>

Explanation:

From the equation of the first law of thermodynamics, ΔU = Q + W

Since there is no expansion work in the bomb calorimeter,  ΔU = Q

But Q = CΔT

where C is heat capacity of the bomb calorimeter =  11.3
kJ/ºC; ΔT = temperature change

For combustion of methane gas:

Q per gram = (
11.3
kJ/ºC * 7.3°C)/1.0g

Q = 83 kJ/g

For combustion of hydrogen gas:

Q per gram = (
11.3
kJ/ºC * 14.3°C)/1.0g

Q = 162 kJ/g

3 0
3 years ago
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