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faust18 [17]
2 years ago
7

Uranium has three common isotopes. If the abundance of 234U is 0.0054%, the abundance of 235U is 0.7204% and the abundance of 23

8U is 99.2742%, what is the average atomic mass of uranium?
Chemistry
1 answer:
OlgaM077 [116]2 years ago
7 0

Answer:

237.9 amu

Explanation:

Given data:

Abundance of U-234 = 0.0054 %

Abundance of  U-235 = 0.7204%

Abundance of  U-238 = 99.2742%

Average atomic mass = ?

Solution:

Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) + (abundance of 3rd isotope × its atomic mass  / 100

Average atomic mass = (0.0054×234)+(0.7204×235) +(99.2742×238)  /100

Average atomic mass =  1.2636+ 169.294+ 23627.2596 / 100

Average atomic mass = 23797.8172/ 100

Average atomic mass = 237.9 amu.

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The receive very little rainfall
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Consider only the experiment you conducted with 0.5 g. of lactose.
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Answer is: <span>the pH value(level) is the independent variable.</span><span>
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3 years ago
The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21 x 10-8 cm. What is this distance in inche
Sergio [31]

Answer:

d=4.75\times 10^{-9}\ \text{inches}

Explanation:

Given that,

The distance between the centers of the two oxygen atoms in an oxygen molecule is 1.21\times 10^{-8}\ cm.

We need to convert this distance in inches.

We know that,

1 cm = 0.393 inches

We can solve it as follows :

1.21\times 10^{-8}\ cm=0.393\times 1.21\times 10^{-8}\\\\=4.75\times 10^{-9}\ \text{inches}

So, the distance between the centers of the two oxygen atoms is 4.75\times 10^{-9}\ \text{inches}.

5 0
2 years ago
A 1.8 g sample of octane C8H18 was burned in a bomb calorimeter and the temperature of 100 g of water increased from 21.36 C to
melomori [17]

Answer:

HEAT OF COMBUSTION PER GRAM OF OCTANE IS 1723.08 J OR 1.72 KJ/G OF HEAT

HEAT OFF COMBUSTION PER MOLE OF OCTANE IS 196.4 KJ/ MOL OF HEAT

Explanation:

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

In other words, 3101.56 J of heat was evolved from the reaction of 1.8 g octane with water.

Heat of combustion of octane per gram:

1.8 g of octane produces 3101.56 J of heat

1 g of octane will produce ( 3101.56 * 1 / 1.8)

= 1723.08 J of heat

So, heat of combustion of octane per gram is 1723.08 J

Heat of combustion per mole:

1.8 g of octane produces 3101.56 J of heat

1 mole of octane will produce X J of heat

1 mole of octane = 114 g/ mol of octane

So we have:

1.8 g of octane = 3101.56 J

114 g of octane = (3101.56 * 114 / 1.8) J of heat

= 196 432.13 J

= 196. 4 kJ of heat

The heat of combustion of octane per mole is 196.4 kJ /mol.

Mass of water = 100 g

Change in temperature = 28.78 °C - 21.36°C = 7.42 °C

Heat capcacity of water = 4.18 J/g°C

Mass of octane = 1.8 g

Molar mass of octane = C8H18 = (12 * 8 + 1 * 18) g/mol= 96 + 18 = 114 g/mol

First is to calculate the heat evolved when 100 g of water is used:

Heat = mass * specific heat capacity * change in temperature

Heat = 100 * 4.18 * 7.42

Heat = 3101.56 J

8 0
2 years ago
What does the enthalpy of reaction measure?
Gnesinka [82]
The heat released/absorbed by a reaction that occurs at constant pressure.

Hope this helped :)
7 0
3 years ago
Read 2 more answers
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