Question #1
Potasium hydroxide (known)
volume used is 25 ml
Molarity (concentration) = 0.150 M
Moles of KOH used
0.150 × 25/1000 = 0.00375 moles
Sulfuric acid (H2SO4)
volume used = 15.0 ml
unknown concentration
The equation for the reaction is
2KOH (aq)+ H2SO4(aq) = K2SO4(aq) + 2H2O(l)
Thus, the Mole ratio of KOH to H2SO4 is 2:1
Therefore, moles of H2SO4 used will be;
0.00375 × 1/2 = 0.001875 moles
Acid (sulfuric acid) concentration
0.001875 moles × 1000/15
= 0.125 M
Question #2
Hydrogen bromide (acid)
Volume used = 30 ml
Concentration is 0.250 M
Moles of HBr used;
0.25 × 30/1000
= 0.0075 moles
Sodium Hydroxide (base)
Volume used 20 ml
Concentration (unknown)
The equation for the reaction is
NaOH + HBr = NaBr + H2O
The mole ratio of NaOH : HBr is 1 : 1
Therefore, moles of NaOH used;
= 0.0075 moles
NaOH concentration will be
= 0.0075 moles × 1000/20
= 0.375 M
Answer:
a) V = 1.542 E11 ft³
b) V = 4367011968 m³
c) V = 1.1535 E12 us gal
Explanation:
Acre surface is defined as 66 by 660 feet at a depth of one foot:
⇒ Vacre-foot = 66ft*660ft*1ft = 43560 ft³
lake:
a) V = 3.54 E6 acre-feet * ( 43560 ft³ / acre-foot ) = 1.542 E11 ft³
b) V = 1.542 E11 ft³ * ( 0.02832m³ / ft³ ) = 4367011968 m³
c) V = 1.542 E11 ft³ * ( 7.48052 us gal/ft³ ) = 1.1535 E12 us gal
9.05 E-22 g. ~ 10.0 E-22 g Cu
Answer:
3.37 m
Explanation:
<u>Number of moles of solute present in 1 kg of solvent is termed as molality</u>
It is represented by 'm'.
Thus,

Given that:
The mass of LiCl = 15.0 g
Molar mass of LiCl = 42.394 g/mol
The formula for the calculation of moles is shown below:
Thus,
Mass of the solvent = 105 g
Also, 1 g = 0.001 g
So,
Mass of water (solvent) = 0.105 kg
Molality is:

<u>Molality = 3.37 m</u>