How many moles of helium atoms are there in

A hypothetical element has an atomic weight of 48.68 amu. It consists of three isotopes having masses of 47.00 amu, 48.00 amu, a

Morgarella [4.7K]

Answer : The percent abundance of the heaviest isotope is, 78 %

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

As we are given that,

Average atomic mass = 48.68 amu

Mass of heaviest-weight isotope = 49.00 amu

Let the percentage abundance of heaviest-weight isotope = x %

Fractional abundance of heaviest-weight isotope = $\frac{x}{100}$

Mass of lightest-weight isotope = 47.00 amu

Percentage abundance of lightest-weight isotope = 10 %

Fractional abundance of lightest-weight isotope = $\frac{10}{100}$

Mass of middle-weight isotope = 48.00 amu

Percentage abundance of middle-weight isotope = [100 - (x + 10)] % = (90 - x) %

Fractional abundance of middle-weight isotope = $\frac{(90-x)}{100}$

Now put all the given values in above formula, we get:

$48.68=[(47.0\times \frac{10}{100})+(48.0\times \frac{(90-x)}{100})+(49.0\times \frac{x}{100})]$

$x=78\%$

Therefore, the percent abundance of the heaviest isotope is, 78 %

5 0

3 years ago

Read 2 more answers

13.8g element consist 4.6 ×10^22 atoms what is atomic mass of element?

Keith_Richards [23]

<h2>~<u>Solution</u> :-</h2>

Here, to find the atomic mass of element, we must;

We know that,

4.6 x $ \sf{10^{22}}$ atoms of an element weigh 13.8g.

Thus,

The atoms of $ \sf{ 6.02 \times 10^{13}}$ will weigh;

$= \sf{\frac{13.8 \times 6.02 \times {10}^{23} }{4.6 \times {10}^{22} } } \\$

$= \sf{180.6 \: g \: \: (molar \: mass)}$

Hence, the molar mass (atomic mass) will be <u>180.6 g.</u>

8 0

2 years ago

Read 2 more answers

When butane burns completely, only water and carbon dioxide gas are produced. If 11.6 g of butane and 40.0 L of oxygen at 22.0o

Angelina_Jolie [31]

19.7 litre volume of carbon dioxide gas at 22.0o C and 102 kPa can be collected over water.

<h3>What is vapour pressure?</h3>

Vapour pressure is a measure of the tendency of a material to change into the gaseous or vapour state, and it increases with temperature.

Moles of Butane = mass in grams / molar mass = 11.6 / 58.12 = 0.2

Volume of $O_2$ (V) = 40 liter

Temperature (T) = 22°C = 22 + 273 = 295 K

Pressure (P) = 102 kPa = 102 / 101.325 = 1.007 atm

Moles of $O_2$ (n) can be calculated by ideal gas equation.

PV = nRT

n = 1.007 40 ÷ 0.0821 295 = 1.663

Balanced chemical reaction;

2 $C_4H_10$ + 13 $O_2$ ---> 8 $CO_2$ + 10 $H_2O$

From reaction;

13 moles $O_2$ require 2 moles $C_4H_10$

So, 1.663 moles $O_2$ will require = 2 x 1.663 ÷13 = 0.256 moles of $C_4H_10$

Thus $C_4H_10$ is a limiting reagent. So it will drive the yield of $CO_2$ .

Moles of $CO_2$ produced = (8/2) 0.2 = 0.8 moles

Pressure of $CO_2$ (P) = 102 - 2.24 = 99.76 kPa = 99.76 ÷ 101.325 = 0.985 atm

Applying the ideal gas equation for $CO_2$ ,

PV = nRT

0.985 V = 0.8 0.0821 x 295

V = 19.7 liter

The volume of $CO_2$ produced = 19.7 liter.

Learn more about the vapour pressure here:

brainly.com/question/25699778

#SPJ1

5 0

1 year ago

A sample of gas occupies 10.0 l at 100.0 torr and 27.0

disa [49]

The pressure of a sample of a gas if the temperature is changed to 127 c while the volume remains constant is calculated using gay lussac law formula

that is P1/T1 = P2/V2
P1 = 100 torr
T1 = 27+273 = 300 k
T2 =127 +273 =400 k
P2=?

by making P2 the subject of the formula
P2=T2P1/T1

=100 x 400/300 = 133.3 torr

5 0

3 years ago

What is the H+ in solution with pH of 10.9

Soloha48 [4]

Answer:

Explanation:

has a pH of 6.6, then what is the H3O+ in solution X? View Answer · What is the pOH of a solution in which (H+)

8 0

3 years ago