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A rock is believed to have been formed 1.25 billion years ago, as calculated by using potassium-40 dating. If the half-life of p

Leya [2.2K]

Answer:

The answer to the question is

50 % of the original amount of potassium 40 will be left after one half life or 1.25 billion years

Explanation:

To solve the question we note that the half life is the time for half of the quantity of substance that undergoes radioactive decay to disintegrate, thus

we have

half life of potassium 40 K₄₀ = 1.25 billion years

To support the believe tht the rock was formed 1.25 billion years ago we have

$N_{(t)} =N_{(0)} (\frac{1}{2}) ^{\frac{t}{t_{\frac{1}{2} } } }$

After 1.25 billion years we have

$N_{(t)} =N_{(0)} (\frac{1}{2}) ^{\frac{1.25billion}{1.25 billion} } } }$ = $N_{(t)} =N_{(0)} (\frac{1}{2}) ^{1 } } }$ =0.5 of $N_{(0)}$ will be left or 50 % of the original amount of potassium 40 will be left

4 0

4 years ago

The heat of vaporization for water is 40. 7 kJ/mol. How much heat energy must 150. 0 g of water absorb to boil away completely?

netineya [11]

The heat absorbed by the water sample for vaporization has been 6,105 kJ. Thus, option D is correct.

The heat of vaporization has been the amount of heat required to vaporize 1 gram of liquid.

The heat required for the vaporization has been given as:

$Q=m\Delta H_{vap}$

<h3 /><h3>Computation for the Heat of vaporization</h3>

The heat of vaporization of water has been given as, $\Delta H_{vap}=40.7\;\rm kJ/mol$

The mass of water sample has been, $m=150\;\rm g$

Substituting the values for the heat energy absorbed, <em>Q:</em>

<em />

<em /> $Q=150\;\times\;40.7\;\rm kJ\\
\textit Q=6,105\;kJ$

The heat absorbed by the water sample for vaporization has been 6,105 kJ. Thus, option D is correct.

Learn more about heat of vaporization, here:

brainly.com/question/2427061

3 0

3 years ago

How do you find percent composition?

Tatiana [17]

Answer:

vdsfdvdsfdsf

Explanation:

https://newsengin.zendesk.com/hc/vew/community/posts/360056216292-%D9%85%D9%88%D9%82%D8%B9-%D8%A7%D9%84%D9%86%D9%88%D8%B1-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D8%A7%D9%84%D8%AB%D8%A7%D9%85%D9%86%D8%A9-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-%D9%82%D9%8A%D8%A7%D9%85%D8%A9-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9

https://newsengin.zendesk.com/hc/pst/community/posts/360056216252-Serial-TV-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-7-%D8%A7%D9%84%D8%B3%D8%A7%D8%A8%D8%B9%D8%A9-%D8%A7%D9%88%D9%86%D9%84%D8%A7%D9%8A%D9%86-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-7-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-HD

https://newsengin.zendesk.com/hc/vew/community/posts/360056216292-%D9%85%D9%88%D9%82%D8%B9-%D8%A7%D9%84%D9%86%D9%88%D8%B1-%D9%85%D8%B3%D9%84%D8%B3%D9%84-%D8%A7%D9%84%D9%85%D8%A4%D8%B3%D8%B3-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D8%A7%D9%84%D8%AB%D8%A7%D9%85%D9%86%D8%A9-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9-%D9%82%D9%8A%D8%A7%D9%85%D8%A9-%D8%B9%D8%AB%D9%85%D8%A7%D9%86-%D8%A7%D9%84%D8%AD%D9%84%D9%82%D8%A9-8-%D9%83%D8%A7%D9%85%D9%84%D8%A9-%D9%85%D8%AA%D8%B1%D8%AC%D9%85%D8%A9

4 0

3 years ago

How many moles of Ca[OH]2 are present in 80.0 g of Ca[OH]2?

artcher [175]

Answer:

1.08mol

Explanation:

moles = reacting mass/ molecular weight

reacting mass = 80.0g molecular weight of Ca[OH]2= 40 + 2(16 +1) = 74g/mol

mole = 80.0/74 = 1.08mol

4 0

4 years ago