Answer:
Sugar can be obtained by evaporation.
Filtering wont work because sugar is soluble in water thus will mix to form one layer.
First, we need to get the value of Ka:
when Ka = Kw / Kb
we have Kb = 1.8 x 10^-5
and Kw = 3.99 x 10^-16 so, by substitution:
Ka = (3.99 x 10^-16) / (1.8 x 10^-5) = 2.2 x 10^-11
by using the ICE table :
NH4+ + H2O →NH3 + H+
intial 0.013 0 0
change -X +X +X
Equ (0.013-X) X X
when Ka = [NH3][H+] / [NH4+]
by substitution:
2.2 x 10^-11 = X^2 / (0.013 - X) by solving this equation for X
∴X = 5.35 x 10^-7
∴[H+] = X = 5.35 x 10^-7
∴PH = - ㏒[H+]
= -㏒(5.35 x 10^-7)
= 6.27
Explanation:
Molar mass
The mass present in one mole of a specific species .
The molar mass of a compound , can easily be calculated as the sum of the all the individual atom multiplied by the number of total atoms .
(a) P₄
Molar mass of of the atoms are -
Phosphorous , P = 31 g/mol
Molecular mass of P₄ = ( 4 * 31 ) = 124 g/mol .
(b) H₂O
Molar mass of of the atoms are -
Hydrogen , H = 1 g/mol
oxygen , O = 16 g/mol.
Molecular mass of H₂O = ( 2 * 1 ) + ( 1 * 16 ) = 18 g/mol
(c) Ca(NO₃)₂
Molar mass of of the atoms are -
calcium , Ca = 40 g/mol
nitrogen, N = 14 g/mol
oxygen , O = 16 g/mol.
Molecular mass of Ca(NO₃)₂ = ( 1 * 40 ) + ( 2 * 14 ) + ( 6 * 16 ) = 164 g/mol.
(d)CH₃CO₂H (acetic acid)
Molar mass of of the atoms are -
Carbon , C = 12 g/mol.
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molecular mass of CH₃CO₂H =( 2 * 12 ) + (2 * 16 ) + (4 * 1 ) = 60 g/mol.
(e) C₁₂H₂₂O₁₁ (sucrose, cane sugar).
Molar mass of of the atoms are -
Carbon , C = 12 g/mol.
oxygen , O = 16 g/mol.
Hydrogen , H = 1 g/mol
Molecular mass of C₁₂H₂₂O₁₁ = (12 * 12 ) + ( 22 * 1 ) + ( 11 * 16 ) = 342 g/mol.
Answer:
7178.22 mL
Explanation:
Applying,
PV = P'V'................. Equation 1
Where P = Initial pressureof gas, V = Initial volume of gas, P' = Final pressure of gas, V' = Final volume of gas.
make V' the subject of the equation
V' = PV/P'.............. Equation 2
From the question,
Given: V = 2250 mL, P = 290 kPa, P' = 0.9 atm = (101×0.9) = 90.9 kPa
Substitute these values into equation 2
V' = (2250×290)/90.9
V' = 7178.22 mL
