CaCO3=100g
CaO=56g
56g of CaO=100g of CaCO3
1.12g of CaO= 100/56*1.12 g of CaCO3
=2g of CaCO3 (ans)
Answer:
B = - 0.0326 dm³/mol
Explanation:
virial eq until second term:
∴ P = 10 bar * (atm/ 1.01325 bar) = 9.869 atm
∴ T = 200°C = 473 K
∴ Vm = 3.90 dm³/mol
∴ R = 0.08206 dm³.atm/K.mol
⇒ PVm / RT = 1 + B/Vm
⇒ ((9.869 atm)*(3.90 dm³/mol)) / ((0.08206 dm³.atm/mol.K)*(473K)) = 1 + B/Vm
⇒ 0.99164 = 1 + B/Vm
⇒ B/Vm = - 8.357 E-3
⇒ B = (3.90 dm³/mol)*( - 8.357 E-3 )
⇒ B = - 0.0326 dm³/mol
Answer:
We need 8.11 grams of glucose for this solution
Explanation:
Step 1: Data given
Molarity of the glucose solution = 0.300 M
Total volume = 0.150 L
The molecular weight of glucose = 180.16 g/mol
Step 2: Calculate moles of glucose in the solution
Moles glucose = molarity solution * volume
Moles glucose = 0.300 M * 0.150 L
Moles glucose = 0.045 moles glucose
Step 3: Calculate mass of glucose
MAss glucose = moles glucose* molecular weight of glucose
MAss glucose = 0.045 moles * 180.16 g/mol
MAss glucose = 8.11 grams
We need 8.11 grams of glucose for this solution
That an ip leak don’t trust it ^^^^^^^^^^
Bronsted lowry bases
NO2- amd OH-
