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:
c) 2.5 mL
Explanation:
Solution
Doctors order = 0.125g
and
The liquid suspension concentration = 250 mg/5ml
= 0.250g/5ml
Or 0.05g/ml
Amount of ml of suspension required = 0.125g/(0.05g/ml) = 2.5ml
The balanced chemical reaction:
C3H8 + 5O2 = 3CO2 + 4H2O
We are given the amount of the carbon dioxide to be produced. This will be the starting point of our calculations.
<span>43.62 L CO2 ( 1 mol CO2 / 22.4 L CO2 ) (5 mol O2 / 3 mol CO2 ) (
22.4 L O2 / 1 mol O2) = 72.7 L O2</span>
Answer:
Magnesium chloride and water
Explanation:
Mg(OH)₂ + 2HCl ⟶ MgCl₂ + 2H₂O
magnesium chloride water
