To calculate the average mass of the element, we take the summation of the product of the isotope and the percent abundance. In this case, the equation becomes 186.207=187*0.626+185*x where x is the percent abundance of 185. The answer is 0.374 or 37.4%. This can also be obtained by 100%-62.6%= 37.4%.
Answer:
0.56 liters
Explanation:
First we <u>convert 0.80 grams of O₂ into moles</u>, using its molar mass:
- 0.80 g ÷ 32 g/mol = 0.025 mol
At STP, 1 mol of any given mass occupies 22.4 L. With that information in mind we <u>calculate the volume that 0.025 moles of O₂ gas would occupy</u>:
- 0.025 mol * 22.4 L/mol = 0.56 L
Thus the answer is 0.56 liters.
☃️ Chemical formulae ➝ 
<h3>
<u>How to find?</u></h3>
For solving this question, We need to know how to find moles of solution or any substance if a certain weight is given.
<h3>
<u>Solution:</u></h3>
Atomic weight of elements:
Ca = 40
C = 12
O = 16
❍ Molecular weight of
= 40 + 12 + 3 × 16
= 52 + 48
= 100 g/mol
❍ Given weight: 10 g
Then, no. of moles,
⇛ No. of moles = 10 g / 100 g mol‐¹
⇛ No. of moles = 0.1 moles
☄ No. of moles of Calcium carbonate in that substance = <u>0.1 moles</u>
<u>━━━━━━━━━━━━━━━━━━━━</u>
The limiting reactant when 5.6 moles of aluminium react with 6.2 moles of water is
water( H2O)
<u><em>Explanation</em></u>
The balanced equation is as below
2 Al +3 H2O → Al2O3 +3 H2
The mole ratio of Al :Al2O3 is 2:1 therefore the moles of Al2O3
= 5.6 x1/2 = 2.8 moles
The mole ratio of H2O: Al2O3 is 3:1 therefore the moles of Al2O3 produced
= 6.2 x1/3= 2.067 moles
since H2O yield less amount of Al2O3 , H2O is the limiting reagent.
