Answer:
57.2 g
Explanation:
First we <u>convert 66.4 grams of Ba(ClO₄)₂·3H₂O into moles</u>, using its <em>molar mass</em>:
- Molar mass of Ba(ClO₄)₂·3H₂O = Molar mass of Ba(ClO₄)₂ + (Molar Mass of H₂O)*3
- Molar mass of Ba(ClO₄)₂·3H₂O = 390.23 g/mol
- 66.4 g ÷ 390.23 g/mol = 0.170 mol Ba(ClO₄)₂·3H₂O
0.170 moles of Ba(ClO₄)₂·3H₂O would produce 0.170 moles of 0.170 moles of Ba(ClO₄)₂. Meaning we now <u>convert 0.170 moles of Ba(ClO₄)₂ into grams,</u> using the molar mass of Ba(ClO₄)₂:
- 0.170 mol * 336.23 g/mol = 57.2 g
There are exactly (a). 10.0 and (b). 28.0
Explanation:
Below is an attachment containing the solution.
Answer:
i mean kinda they should throw people in the air tho lol
Explanation:
The volume of CO2 at STP =124.298 L
<h3>Further explanation</h3>
Given
Reaction
4 KMnO4, +4 C3H5(OH)5, -7K2CO3, + 7 Mn2O3, +5 CO2, + 16 H2O
701,52 g of KMnO4
Required
volume of CO2 at STP
Solution
mol KMnO4 (MW=158,034 g/mol) :
mol = mass : MW
mol = 701.52 : 158.034
mol = 4.439
mol CO2 from equation : 5/4 x mol KMnO4 = 5/4 x 4.439 = 5.549
At STP 1 mol = 22.4 L, so for 5.549 moles :
=5.549 x 22.4
=124.298 L