Answer:
The answer would be B, PC13
Explanation:
Answer:
357 g of the transition metal are present in 630 grams of the compound of the transition metal and iodine
Explanation:
In any sample of the compound, the percentage by mass of the transition metal is 56.7%. This means that for a 100 g sample of the compound, 56.7 g is the metal while the remaining mass, 43.3 g is iodine.
Given mass of sample compound = 630 g
Calculating the mass of iodine present involves multiplying the percentage by mass composition of the metal by the mass of the given sample;
56.7 % = 56.7/100 = 0.567
Mass of transition metal = 0.567 * 630 = 357.21 g
Therefore, the mass of the transition metal present in 630 g of the compound is approximately 357 g
Answer:
D
Explanation:
the answer is d white precipitate
Answer:
0.0907 M
Explanation:
Before you can calculate the molarity, you need to convert grams to moles (via molar mass) and convert mL to L.
(Step 1)
Molar Mass (C₈H₅O₄K):
8(12.011 g/mol) + 5(1.008 g/mol) + 4(15.998 g/mol) + 39.098 g/mol
Molar Mass (C₈H₅O₄K): 204.218 g/mol
0.6013 g C₈H₅O₄K 1 mole
------------------------------ x ------------------ = 0.00294 moles C₈H₅O₄K
204.218 g
(Step 2)
1,000 mL = 1 L
32.47 mL 1 L
--------------- x ----------------- = 0.03247 L
1,000 mL
(Step 3)
Molarity (M) = moles / volume (L)
Molarity = 0.00294 moles / 0.03247 L
Molarity = 0.0907 M
