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2 years ago

How much mass is lost through radioactive decay if 1.8 x 1015 Jare.

Chemistry

1 answer:

ser-zykov [4K]2 years ago

7 0

The mass lost through radioactive decay is 0.094 kg.

<h3>What is the mass lost in a radioactive decay?</h3>

The mass loss in a nuclear reaction such as a radioactive decay is known as mass defect.

The lost mass is converted to energy released during the decay process.

The energy and mass are related by the formula:

E = mc²

where

m is mass

E is energy

c is speed of light = 3.0 * 10⁸ m/s

Mass lost = E/c²

mass lost = 8.5 * 10¹⁵/(3.0 * 10⁸)²

mass lost = 0.094 kg

Therefore the mass lost is 0.094 kg.

Learn more about mass defect at: brainly.com/question/16485729

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3 years ago

Sodium chloride (NaCl) is the chemical name for table salt and potassium chloride (KCl) is a common salt substitute. Using the p

Lina20 [59]

Answer:

Option B. Sodium and potassium lose electrons and chlorine gains them

Explanation:

Sodium and Potassium are both group 1 element. They both react with chlorine by losing 1 electron each and the chlorine atom accept the electron.

Sodium chloride(NaCl) can be formed by the equation below:

Na —> Na+ + e-

Cl + e- —> Cl-

Combining both equation together we obtained the following:

Na + Cl + e- —> Na+ + Cl- + e-

Eliminating the e-, we have:

Na + Cl —> Na+Cl-

From the above illustration, we can see that sodium loses electron and chlorine accepts/gain the electron

This is also applicable to potassium chloride as shown below:

K —> K+ + e-

Cl + e- —> Cl-

Combining both equation together we obtained the following:

K + Cl + e- —> K+ + Cl- + e-

Eliminating the e-, we have:

K + Cl —> K+Cl-

we can see also that potassium loses electron and chlorine accepts/gain the electron

Therefore, the correct answer to the question is option B

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4 years ago

What is colloidal solutions <br>

Igoryamba

Explanation:

Colloidal solutions, or colloidal suspensions, are nothing but a mixture in which the substances are regularly suspended in a fluid. ... Colloidal systems can occur in any of the three key states of matter gas, liquid or solid. However, a colloidal solution usually refers to a liquid concoction.

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3 years ago

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Energy can make matter______. A.move. B.change. C.work. D.all of the above.

AysviL [449]

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3 years ago

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Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670

Aliun [14]

Answer : The empirical formula of the compound is $C_3H_8O_2$

Explanation :

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.043g$

Mass of $H_2O=0.5670g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 1.043 g of carbon dioxide, $\frac{12}{44}\times 1.043=0.284g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.5670 g of water, $\frac{2}{18}\times 0.5670=0.063g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0158 moles.

For Carbon = $\frac{0.0237}{0.0158}=1.5$

For Hydrogen = $\frac{0.063}{0.0158}=3.98\approx 4$

For Oxygen = $\frac{0.0158}{0.0158}=1$

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is $C_3H_8O_2$

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3 years ago