Chemistry College

Find the mass in grams of 1.38 moles of Sr

Answers

Answer 1

Answer:

116.78 grams.

Explanation:

1 mol of Strontium (Sr) = 87.62 grams

1.38 mol of Strontium = x

Cross Multiply

1 * x = 1.38 * 87.62

x = 116.78 grams

Related Questions

Use this equation for the next question:
2NaOH + H2SO4 ® Na2SO4 + 2H20
If a reaction produces 0.75 moles Na2SO4, how many moles of NaOH were used?
0.75 moles NaOH
2 moles NaOH
.375 moles NaOH
1.5 moles NaOH

Answers

The answers is: 375 moles NaOH

An element has five isotopes. Calculate the atomic mass of this element using the information below. Show all your work. Using the periodic table, identify the element this is likely to be and explain your choice. (18 pts)
A) Isotope 1 – mass: 64 amu; percent abundance: 48.89%
B) Isotope 2 – mass: 66 amu; percent abundance: 27.81%
C) Isotope 3 – mass: 67 amu; percent abundance: 4.11%
D) Isotope 4 – mass: 68 amu; percent abundance: 18.57%
E) Isotope 5 – mass: 70 amu; percent abundance: 0.62%

Answers

Answer: Sol:-

Data provided in the question is :-

Atomic mass of isotope -1 = 64 amu

Atomic mass of isotope -2 = 66 amu

Atomic mass of isotope -3 = 67 amu

Atomic mass of isotope -4 = 68 amu

Atomic mass of isotope - 5 = 70 amu

Percentage abundace of isotope - 1 = 48.89 %

Percentage abundance of isotope -2 = 27.81 %

Percentage abundance of isotope - 3 = 4.11%

Percentage abundance of isotope-4 = 18.57%

Percentage abundance of isotope - 5 = 0.62 %

Formula used :-

Average atomic mass of an element =[ {(atomic mass of isotope-1 * percentage abundance of isotope-1) + ( atomic mass of isotope-2 * percentage abundance of isotope -2) + ( atomic mass of isotope -3 * percantege abundance of isotope-3 ) + ( atomic mass of isotope-4 * percentage abundance of isotope-4) + (atomic mass of isotope-5 * percentage abundance of isotope-5)} / 100]

Calculation :-

Put all the value in the formula :-

Average atomic mass of an element = [{(64 * 48.89) + (66 * 27.81) + (67 * 4.11) + (68 * 18.57) + (70 * 0.62)} / 100] amu

= [{(3128.96) + (1835.46) +(257.37) + (1262.76) + (43.4)} / 100] amu

= {(6528.04) / 100} amu

= 65.2804 amu

Average atomic mass of an element is = 65.2804 amu

Then this mass is approximatly equal to atomic mass of zinc so this element would be zinc

atomic mass of zinc = 65.38 \approx 65.2804 amu

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