Answer:
C. 7.50g
Explanation:
The percent (%) by mass of a solute in a solution refers to the number of grams contained in 100g of solution by that solute. In this case, 5% by mass of pottasium chloride (KCl) means 5g of KCl is contained in 100g of solution.
Therefore, in 150g of solution, there would be:
5g/100g × 150g
= 0.05 × 150
= 7.50g of KCl solute.
Hence, 7.50g of pottasium chloride would be expected to be collected by evaporating 150.0 g of the solution.
Answer: 11,200 g
step by step explanation:
D= M/V
M= DV
M= (13.6 g/cm^3) (825cm^3)
M= 11,200 g
Increase temperature = particle move faster
increase Concentration =more crowded so more collisions
Explanation:
when temperature increase,the particle gain kinetic energy which make the particles move faster hence more collisions.
if concentration increase,there will be more frequent collisions between the particles due to crowded
Answer:
43.2 moles of carbon dioxide are required and 421g of glucose could be produced
Explanation:
Based on the reaction:
6CO2 + 6H2O → C6H12O6 + 6O2
1 mole of glucose, C6H12O6, requires 6 moles of carbon dioxide. 7.2moles of glucose requires:
7.2mol C6H12O6 * (6mol CO2 / 1mol C6H12O6) =
<h3>43.2 moles of carbon dioxide are required</h3><h3 />
618g of CO2 -Molar mass: 44.01g/mol- are:
618g * (1mol / 44.01g) = 14.04moles CO2
Moles C6H12O6:
14.04moles CO2 * (1mol C6H12O6 / 6mol CO2) = 2.34moles C6H12O6
Mass glucose -Molar mass: 180.156g/mol-
2.34moles C6H12O6 * (180.156g / mol) =
<h3>421g of glucose could be produced</h3>