First, We have to write the equation for neutralization:
Ba(OH)2 + 2HCl → BaCl2 + 2H2O
so, from the equation of neutralization, we can get the ratio between Ba(OH)2 and HCl. Ba(OH)2 : HCl = 1:2
- We have to get the no.of moles of Ba(OH)2 to do the neutralization as we have 25.9ml of 3.4 x 10^-3 M Ba(OH)2.
So no.of moles of Ba(OH)2 = (25.9ml/1000) * 3.4x10^-3 = 8.8 x 10^-5 mol
and when Ba(OH)2 : HCl = 1: 2
So the no.of moles of HCl = 2 * ( 8.8x10^-5) = 1.76 x 10^-4 mol
So when we have 1.76X10^-4 Mol in 16.6 ml (and we need to get it per liter)
∴ the molarity = no.of moles / mass weight
= (1.76 x 10^-4 / 16.6ml)* (1000ml/L) = 0.0106 M Hcl
Answer:
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Explanation: I think this is what you are looking for. Hope this helps.
Answer:
124 g (3 sig figs)
or
124.011 g (6 sig figs
Explanation:
Step 1: Calculate g/mol for AgNO₃
Ag - 107.868 g/mol
N - 14.01 g/mol
O - 16.00 g/mol
107.868 + 14.01 + 16.00(3) = 169.878 g/mol
Step 2: Multiply 0.73 moles by molar mass
0.73 mol (169.979 g/mol)
124 grams of AgNO₃
Given data:
Mass of cofactor A in an average yeast cell = 96.15 pg
Now:
1 picogram (pg) = 1*
g
1 microgram (ug) = 1*
g
Therefore, 1 pg = 1*ug
Mass of cofactor A in one yeast cell = 96.15 * micro gram
Number of cells in the yeast colony = 105
Therefore, the total mass of cofactor A is given as:
Number of cells * mass of cofactor A per cell
= 105 cells * 96.15 * micro grams/cell = 1.009 * 
micro grams
