Oxygen is used up and glucose is broken down.
Answer:
Option A. FeCl3
Explanation:
The following data were obtained from the question:
Mass of iron (Fe) = 6.25g
Mass of the compound formed = 18g
From the question, we were told that the compound formed contains chlorine. Therefore the mass of chlorine is obtained as follow
Mass of chlorine (Cl) = Mass of compound formed – Mass of iron.
Mass of chlorine (Cl) = 18 – 6.25
Mass of chlorine (Cl) = 11.75g
The compound therefore contains:
Iron (Fe) = 6.25g
Chlorine (Cl) = 11.75g
The empirical formula for the compound can be obtained by doing the following:
Step 1:
Divide by their molar mass
Fe = 6.25/56 = 0.112
Cl = 11.75/35.5 = 0.331
Step 2:
Divide by the smallest
Fe = 0.112/0.112 = 1
Cl = 0.331/0.112 = 3
The empirical formula for the compound is FeCl3
Hey There!
At neutralisation moles of H⁺ from HCl = moles of OH⁻ from Ca(OH)2 so :
0.204 * 42.8 / 1000 => 0.0087312 moles
Moles of Ca(OH)2 :
2 HCl + Ca(OH)2 = CaCl2 + 2 H2O
0.0087312 / 2 => 0.0043656 moles ( since each Ca(OH)2 ives 2 OH⁻ ions )
Therefore:
Molar mass Ca(OH)2 = 74.1 g/mol
mass = moles of Ca(OH)2 * molar mass
mass = 0.0043656 * 74.1
mass = 0.32 g of Ca(OH)2
Hope that helps!
Answer:
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
Explanation:
Using the formula
Ca Va = Cb Vb
Cb = 0.32 M
Vb = 50 mL = 50/1000 = 0.050L
Ca = 0.5 M
Va =?
Substituting for Va in the equation, we obtain:
Va = Cb Vb / Ca
Va = 0.32 * 0.05 / 0.5
Va = 0.016 / 0.5
Va = 0.032 L
The volume of HCl to be added to completely react with the ammonia is 0.032 L or 32mL
