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Alik [6]
1 year ago
12

Use the atomic mass of indium to calculate the relative abundance of indium-113.

Chemistry
1 answer:
ASHA 777 [7]1 year ago
7 0

The relative abundance of indium-113 is 4%.

The isotopes are species of the same element having the same atomic number but a different mass number.

The elements occurring in nature exist as multiple isotopes.

When we take into account the existence of these isotopes and their relative abundance (percent), the average atomic mass of that element can be computed, which is given by the following formula,

Average atomic Mass= (%age of isotope 1) x (Mass of isotope 1) + (%age of isotope 2) x (Mass of isotope 2)/100

Indium exists in the form of Indium-113 and Indium-115.

The mass of Indium-113 is 112.90 u.

The mass of Indium-115 is 114.90 u.

The average atomic mass of Indium is 114.82 u.

Let the %age of isotope 1(Indium-113) be X.

Then, the %age of isotope 2(Indium-115) would be 100-X.

Applying the values in the formula,

Average atomic mass = 112.90X+114.90(100-X)

114.82 = 112.90X+114.90(100-X)

On solving the above equation, the value of X comes out to be 4%.

Thus, the relative abundance/%age abundance of Indium-113 is 4%.

To know more about "Average Atomic Mass", refer to the following link:

brainly.com/question/13753702?referrer=searchResults

#SPJ4

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djverab [1.8K]

Answer:

29.92grams of PbSO4

Explanation:

lead (iV) oxide = PbO2 = Molar mass: 239.2 g/mol

lead (ll) sulfate = PbSO4 = Molar mass: 303.26 g/mol

PbO2 = PbSO4

1:1 ratio

Pb = Lead

Lead has an oxidation number of 4+

O = Oxygen

Oxygen has an oxidation number of 2-

PbO2 + 4H+ + SO4 2- + 2e- = PbSO4(s) + 2H2O

Ok so the above would be the likely complete reaction, though we don't really need this as we already know the ratio is 1:1.

23.6g of PbO2

23.6/239.2 = 0.09866 Moles of PbO2

Since we have a 1:1 ratio we know that the same number of moles of PbSO4 are produced and since we know the molar mass it's simply molar mass multiplied by number of moles.

303.26 x 0.09866 = 29.92grams of PbSO4

6 0
3 years ago
In oxidative phosphorylation, electrons are passed from one electron carrier to another. The energy released is used to ________
Zepler [3.9K]

Answer:

d. Form ATP during glycolysis.

Explanation:

During oxidative phosphorylation, an oxidation-reduction reactions will happen and the electrons transferred and energy released will be used in the conversion of adenosine diphosphate (ADP) into adenosine triphosphate (ATP).

This happens as a step of glycolysis, wich is the process in which the organism breaks gluscose molecules to obtain energy.

7 0
4 years ago
). In a titration, a student obtained an average titre value of 3.9 cm3 of 0.3 M HCl. If the volume of Na2CO3 solution used is 1
Oliga [24]

Answer:

I) 0.0585 M

ii)6.2 g dm-3

Explanation:

The reaction equation is given as;

Na2CO3(aq) +2HCl(aq)------> 2NaCl(aq) + CO2(g) +H2O(l)

Concentration of acid CA= 0.3 M

Volume of acid VA= 3.9 cm^3

Concentration of base CB= the unknown

Volume of base VB= 10 cm^3

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB=CBVBNA

CB= CAVANB/VBNA

substituting values;

CB= 0.3 × 3.9 × 1/ 10.0 × 2

CB= 0.0585 M

ii) mass concentration= molar concentration × molar mass

Molar mass of Na2CO3= 106 gmol-1

Mass concentration= 0.0585 × 106 = 6.2 g dm-3

5 0
3 years ago
When oxygen and calcium form an ionic bond, what is the formula? o2ca2 oca ca2o2 cao
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6 0
3 years ago
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Which statement describes the formation of metamorphic rocks?
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