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wariber [46]
3 years ago
8

A sample of element X contains 90% X-35 atoms, 8.0% X-37 atoms, and 2.0% X-38 atoms. The average atomic mass will be closest to

which value?
Chemistry
2 answers:
Ne4ueva [31]3 years ago
6 0

To find average atomic mass you multiply the mass of each isotope by its percentage, and then add the values up.

35 * 0.90 + 37 * 0.08 + 38 * 0.02 = 35.22

Average atomic mass closest to 35.22 amu.

brilliants [131]3 years ago
4 0

Answer : The average atomic mass will be closest to 35 amu.

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,

Mass of isotope X-35 = 35 amu

Percentage abundance of isotope X-35 = 90 %

Fractional abundance of isotope X-35 = 0.90

Mass of isotope X-37 = 37 amu

Percentage abundance of isotope X-37 = 8.0  %

Fractional abundance of isotope X-37 = 0.08

Mass of isotope X-38 = 38 amu

Percentage abundance of isotope X-38 = 2.0  %

Fractional abundance of isotope X-38 = 0.02

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

\text{Average atomic mass of element}=\sum[(35\times 0.90)+(37\times 0.08)+(38\times 0.02)]

\text{Average atomic mass of element X}=35.22amu\approx 35amu

Therefore, the average atomic mass will be closest to 35 amu.

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The relative atomic mass of Chlorine is 35.45. Calculate the percentage abundance of the two isotopes of Chlorine, 35Cl and 37Cl
nata0808 [166]

Answer:

35Cl = 75.9 %

37Cl = 24.1 %

Explanation:

Step 1: Data given

The relative atomic mass of Chlorine = 35.45 amu

Mass of the isotopes:

35Cl = 34.96885269 amu

37Cl = 36.96590258 amu

Step 2: Calculate percentage abundance

35.45 = x*34.96885269 + y*36.96590258

x+y = 1  x = 1-y

35.45 = (1-y)*34.96885269 + y*36.96590258

35.45 = 34.96885269 - 34.96885269y +36.96590258y

0.48114731 = 1,99704989‬y

y = 0.241 = 24.1 %

35Cl = 34.96885269 amu = 75.9 %

37Cl = 36.96590258 amu = 24.1 %

3 0
2 years ago
35.2 J of heat is
ruslelena [56]

The specific heat : c = 0.306 J/g K

<h3>Further explanation</h3>

Given

Heat = 35.2 J

Mass = 16 g

Temperature difference : 7.2 K =

Required

The specific heat

Solution

Heat can be calculated using the formula:  

Q = mc∆T  

Q = heat, J  

m = mass, g  

c = specific heat, joules / g ° C  

∆T = temperature difference, ° C / K  

Input the value :

c = Q / m.∆T  

c = 35.2 / 16 x 7.2

c = 0.306 J/g K

7 0
3 years ago
What is the molarity of a solution containing 55.8 g of mgcl2 dissolved in 1.00 l of solution?
MakcuM [25]
The answer is 0.59 M.

Molar mass (Mr) of MgCl₂ is the sum of the molar masses of its elements.
So, from the periodic table:
Mr(Mg) = 24.3 g/l
Mr(Cl) = 35.45 g/l
Mr(MgCl₂) = Mr(Mg) + 2Mr(Cl) = 24.3 + 2 · 35.45 = 24.3 + 70.9 = 95.2 g/l

So, 1 mol has 95.2 g/l.

Our solution contains 55.8g in 1 l  of solution, which is 55.8 g/l

Now, we need to make a proportion:
1 mole has 95.2 g/l, how much moles will have 55.8 g/l:
1 M : 95.2 g/l = x : 55.8 g/l
x = 1 M · 55.8 g/l ÷ 95.2 g/l ≈ 0.59 M
7 0
3 years ago
Read 2 more answers
A 35.40 gram hydrate of sodium carbonate, Na2CO3•nH2O, is heated to a constant mass. Its final weight is 30.2 g. What is formula
Sever21 [200]

Answer:

Na₂CO₃•H₂O

Explanation:

After it is heated, the remaining mass is the mass of sodium carbonate.

30.2 g Na₂CO₃

Mass is conserved, so the difference is the mass of the water:

35.4 g − 30.2 g = 5.2 g H₂O

Convert masses to moles:

30.2 g Na₂CO₃ × (1 mol Na₂CO₃ / 106 g Na₂CO₃) = 0.285 mol Na₂CO₃

5.2 g H₂O × (1 mol H₂O / 18.0 g H₂O) = 0.289 mol H₂O

Normalize by dividing by the smallest:

0.285 / 0.285 = 1.00 mol Na₂CO₃

0.289 / 0.285 = 1.01 mol H₂O

The ratio is approximately 1:1.  So the formula of the hydrate is Na₂CO₃•H₂O.

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2 years ago
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Answer:

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