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1 year ago

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A 200. 0-ng sample of sodium-24 was stored in a lead-lined cabinet for 45 hours. The half-life of sodium-24 is 15 hours. How muc

h sodium-24 remains?

1 answer:

ipn [44]1 year ago

3 0

Answer:

25 gm would be left

Explanation:

45 hours is 3 half lives

200 gm * 1/2 * 1/2 * 1/2 = 25 gm

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mixas84 [53]

Answer:

Group 18th

Explanation:

The group 18th of periodic table which is composed of Noble gases mainly was missing from the Mendeleev's periodic table. The reason for this is that Noble gases are least reactive, therefore, it was difficult to predict such elements which do not react with other elements.

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3 years ago

Why is the metric system the preferred system of measurement for science

Nikolay [14]

In is a lot easier because it uses 10s.

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3 years ago

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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:

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2 years ago

What is the entropy change in the environment when 5.0 MJ of energy is transferred thermally from a reservoir at 1000 K to one a

Leni [432]

Answer:

The entropy change in the environment is 3.62x10²⁶.

Explanation:

The entropy change can be calculated using the following equation:

$\Delta S = \frac{Q}{k_{B}}(\frac{1}{T_{f}} - \frac{1}{T_{i}})$

Where:

Q: is the energy transferred = 5.0 MJ

$k_{B}$ : is the Boltzmann constant = 1.38x10⁻²³ J/K

$T_{i}$ : is the initial temperature = 1000 K

$T_{f}$ : is the final temperature = 500 K

Hence, the entropy change is:

$\Delta S = \frac{5.0 \cdot 10^{6} J}{1.38 \cdot 10^{-23} J/K}(\frac{1}{500 K} - \frac{1}{1000 K}) = 3.62 \cdot 10^{26}$

Therefore, the entropy change in the environment is 3.62x10²⁶.

I hope it helps you!

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3 years ago

hich of the following is TRUE? Group of answer choices A neutral solution does not contain any H3O+ or OH- A neutral solution co

Studentka2010 [4]

Answer:

a basic solution has [OH-] > [H3O+]

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3 years ago