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Dvinal [7]
3 years ago
6

1. Radioactive decay of 40K atoms in an igneous rock has resulted in a ratio of 25 percent 40K atoms to 75 percent 40Ar and 40Ca

atoms. How many years old is this rock?
Chemistry
1 answer:
zhuklara [117]3 years ago
5 0
1/2 of the original parent has decayed, thus one half- life has passed, and the rock is 1.3 billion = 1300 million years old
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What are some of the main ways to add energy to a reaction? plzzz help
Tresset [83]

Answer:

Heat.

Light.

Sound.

Electricity

Explanation:

4 0
2 years ago
What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100
Kisachek [45]

What is the percent by mass of sodium in Na2SO4? total mass of element in compound molar mass of compound Use %Element x 100

5 0
3 years ago
How many grams of n2f4 can be produced from 225g f2?
3241004551 [841]
The answer is 615.91 grams of <span>n2f4

Solution:
225g F2 x [(1molF2)/(38gramsF2)] x [</span>(1molF2)/(1molN2F4)] x [(104.02 grams N2F4)/(1molN2F4)]
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8 0
3 years ago
The [-oh) in a sample of egg whites is 6.3 x 10-7 m. what is the [h3o+] in these<br>egg whites?
Kamila [148]

The hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.

<h3>How to calculate [H3O+]?</h3>

The hydrogen ion concentration of a solution can be calculated as follows:

pOH = - log [OH-]

pOH = - log [6.3 × 10-⁷M]

pOH = - [-6.2]

pOH = 6.2

Since pOH + pH = 14

pH = 14 - 6.2

pH = 7.8

pH = - log [H3O+]

7.8 = - log [H3O+]

[H3O+] = 10-⁷:⁸

[H3O+] = 1.5 × 10-⁸M

Therefore, the hydrogen ion concentration [H3O+] in an egg white containing 6.3 × 10-⁷M of [OH-] is 1.5 × 10-⁸M.

Learn more about hydrogen ion concentration at: brainly.com/question/15082545

7 0
1 year ago
Calcium oxide or quicklime (CaO) is used in steelmaking, cement manufacture, and pollution control. It is prepared by the therma
Elena-2011 [213]

Answer:

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

Explanation:

CaCO_3(s)\rightarrow CaO(s) + CO_2(g)

Annual production of CaO = 8.6\times 10^{10} kg=8.6\times 10^{13} g

Moles of CaO :

\frac{8.6\times 10^{13} g}{56 g/mol}=1.53\times 10^{12} moles

According to reaction, 1 mole of CaO is produced along with 1 mole of carbon-dioxide.

Then along with  1.53\times 10^{12} moles of CaO moles of carbon-dioxide moles produced will be:

\frac{1}{1}\times 1.53\times 10^{12} moles=1.53\times 10^{12} moles of carbon-dioxide

Mass of 1.53\times 10^{12} moles of carbon-dioxide:

1.53\times 10^{12}mol\times 44 g/mol=6.73\times 10^{13} g =6.73\times 10^{10} kg

The yearly release of CO_2 into the atmosphere is 6.73\times 10^{10} kg.

6 0
3 years ago
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