Answer:
a. 1 mole of raindrops weights 18 g
b. 5,4x10²¹ moles of raindrops are in Pacific Ocean
Explanation:
a. Chemical composition of raindrops is water that is composed by two hydrogen (One hydrogen mole weights 1 g) and one oxygen (One oxygen mole weights 16 g). Thus, one water mole weights:
16 g + 1 g + 1 g = 18 g
b. 96,5 % of Pacific Ocean is water (Raindrops). So, the mass of raindrops in Pacific Ocean is:
1x10²⁰ kg × 96,5% = 9,65×10¹⁹ kg of raindrops
With the molar mass of raindrops (18 g/ 1 mol) it is possible to know the moles of raindrops in the Pacific Ocean, thus:
9,65×10¹⁹ kg of raindrops × (1000 g / 1 kg) × ( 1 mol of raindrops / 18 g) =
kg to g g to moles
5,4x10²¹ moles of raindrops in the Pacific Ocean
I hope it helps!
Explanation:
a) 1 mole = [tex]6.022\times 10^{23} [/tex] particles/atoms/ molecules
Mass of a rain drop = 50 mg
Mass of [tex]6.022\times 10^{23} [/tex] rain drops:
[tex]6.022\times 10^{23} \times 50 mg=3.011\times 10^{25} mg\approx 3.0\times 10^{25} mg[/tex]
Mass of 1 mole of rain drop = [tex]3.0\times 10^{25} mg[/tex]
b) Mass of pacific ocean = M = [tex]7.08\times 10^{20} kg[/tex]
1 kg = 1,000,000 mg
[tex]M=7.08\times 10^{20} kg =7.08\times 10^{20} \times 10^6 mg[/tex]
[tex]M=7.08\times 10^{26} mg[/tex]
Mass of 1 mole of rain drop = [tex]3.0\times 10^{25} mg[/tex]
Let the number moles rain drops in the pacific ocean be x.
[tex]x\times 50 mg=M[/tex]
[tex]x=\frac{M}{50 mg}[/tex]
[tex]=\frac{7.08\times 10^{26} mg}{3.0\times 10^{25} mg}=23.6 [/tex]
23.6 ≈ 24
There are 24 moles of rain drops in the pacific ocean.
