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wariber [46]
4 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]4 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]4 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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Dmitriy789 [7]

Answer:

option A = S(s) + O₂(g)   →   SO₂ (s)

Explanation:

Chemical equation:

S(s) + O₂(g)   →   SO₂ (s)

when sulfur burned in the presence of oxygen it produce sulfur dioxide. The sulfur dioxide can further react with oxygen to produce sulfur trioxide and then react with water to form sulfuric acid.

Uses of sulfur dioxde:

It is used as a solvent and reagent in laboratory.

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7 0
3 years ago
An excess of Ba(No3)2 reacts with 250ml of H2SO4 solution to give 0.55g of BaSo4.determine The concentration in moles per litre
Wewaii [24]
Chemical reaction: Ba(NO₃)₂ + H₂SO₄ → BaSO₄ + 2HNO₃.
V(H₂SO₄) = 250 mL ÷ 1000 mL/L = 0,25 L.
m(BaSO₄) = 0,55 g.
n(BaSO₄) = m(BaSO₄) ÷ M(BaSO₄).
n(BaSO₄) = 0,55 g ÷ 233,38 g/mol.
n(BaSO₄) = 0,00235 mol.
From chemical reaction: n(BaSO₄) : n(Ba(NO₃)₂) = 1 : 1.
n(Ba(NO₃)₂) = 0,00235 mol.
c(Ba(NO₃)₂) = n(Ba(NO₃)₂) ÷ V.
c(Ba(NO₃)₂) = 0,00235 mol ÷ 0,25 L.
c(Ba(NO₃)₂) = 0,0095 mol/L.
7 0
3 years ago
If there are 3.10 moles of O, how many moles of each of the compounds are present?
Burka [1]

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The question pretty  much requires us to find the amount of moles of each compounds based on the number of moles of O given.

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x mol of H2SO4 would contain 3.10 mol of O

x = 3.10 * 1 / 4 = 0.775 mol of H2SO4

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1 mol of C2H4O2 contains 2 mol of O

x mol of C2H4O2 would contain 3.10 mol of O

x = 3.10 * 1 / 2 = 1.55 mol of C2H4O2

NaOH

1 mol of NaOH contains 1 mol of O

x mol of NaOH would contain 3.10 mol of O

x = 3.10 * 1 / 1 = 3.10 mol of NaOH

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