The balanced reaction is 2KClO3 --> 2KCl + 3O2
We first divide the 400.0 g KClO3 by the molar mass of 122.55 g/mol to get 3.26 mol KClO3. Next, we use the coefficients: 3.26 mol KClO3 * (3 mol O2 / 2 mol KClO3) = 4.896 mol O2. Multiplying this by the molar mass of 32 g/mol gives 156.67 g O2.
Percent yield = 115.0 g / 156.67 g = 0.734 = 73.40%
The amount remaining at the end of 5 half-lives is 7.81×10¹³ g
From the question given above, the following data were obtained:
- Half-life (t½) = 5730 years
- Original amount (N₀) = 2.5×10¹⁵ g
- Number of half-lives (n) = 5
- Amount remaining (N) =?
The amount remaining can be obtained as follow:
N = 1/2ⁿ × N₀
N = 1/2⁵ × 2.5×10¹⁵
N = 1/32 × 2.5×10¹⁵
N = 0.03125 × 2.5×10¹⁵
N = 7.81×10¹³ g
Therefore, the amount remaining after 5 half-lives is 7.81×10¹³ g
Learn more about half-life: brainly.com/question/25783920
Answer:
Moles de LiOH= 0.33212 Moles
Explanation:
1 Mol of LiOH -->6.022*10^23 molecules of LiOH
therefore
Explanation:
Molar mass of O2: 32g/mol
Moles of O2 = 19.5 / 32 = 0.609mol
Answer:
The total water content in the earth remains nearly constant.
Explanation:
- The amount of water on earth is nearly same as the amount of water that was on earth in Mesozoic era.
- Plants and animals use water as a raw material in their growth, but that water is eventually returned to earth.
- When the UV light breaks the water molecules in higher atmosphere, some water is lost.But it is compensated by the water that enters the earth through meteors and comets.
- The water content is not constant but through the series of biochemical reactions,the water level is nearly the same.
