D. several because poly means many or lots
Answer:
103.00
Explanation:
1 mole of Glucose
6 C = 6 * 12 = 72 grams
12H = 12 * 1 = 12 grams
6O = 6 * 16 =96 grams
Total = 180 grams
0.572 moles of Glucose
1 mol of glucose = 180 grams
0.572 mols of glucose = x
x = 0.572 * 180
x = 103.00 grams
Hence the resulting concentration of Li+ is 1M. Hope it helps.
Answer:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
Explanation:
The equilibrium of sodium acetate is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵
Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.
For [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:
[CH₃COO⁻] = 0,1 M - x
[H⁺] = 0,0025 M - x
[CH₃COOH] = x
The expression for this equilibrium is:
Ka =
Replacing:
1,8x10⁻⁵ =
Thus:
0 = x²-0,102518x +2,5x10⁻⁴
Solving:
x = 0,100 ⇒ No physical sense
x = 0,0024995
Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷
pH = - log [H⁺] = 6,30
Following the same procedure changing both [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
I hope it helps!
Answer:
The limiting reacting is O2
Explanation:
Step 1: data given
Number of moles O2 = 21 moles
Number of moles C6H6O = 4.0 moles
Step 2: The balanced equation
C6H6O + 7O2 → 6CO2 + 3H2O
Step 3: Calculate the limiting reactant
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
O2 is the limiting reactant. It will completely be consumed (21 moles).
C6H6O is in excess.
For 7 moles O2 we need 1 mol C6H6O
For 21 moles O2 we'll need 21/7 = 3 moles C6H6O
There will remain 4.0 - 3.0 = 1 mol C6H6O
Step 4: calculate products
For 1 mol C6H6O we need 7 moles O2 to produce 6 moles CO2 and 3 moles H2O
For 21 moles O2 we'll have 6/7 * 21 = 18 moles CO2
For 21 moles O2 we'll have 3/7 * 21 = 9 moles H2O
The limiting reacting is O2