Answer:
a) HNO3 -> H+ + NO3- disassociation of Nitric Acid; to yield a Nitrate ion and a Proton, H+, or as a Hydronium ion H3O+
b) H2S04 -> Disassociation of Sulfuric Acid; simple way- 2H+ + SO4- -
c) H2S hydrogen sulphide in water is an acid; thus H+ HS- disassociation.
d) NaOH -> dissociation of Na+ + OH-; this is complete; sodium hydroxide is deliquescent, meaning it will draw water - EVEN from the air! Strong Base
e) Na2CO3 -> 2Na+ CO3- - Ionization of sodium carbonate - a salt
f) Na2S04 -> 2Na+ + SO4 - - ionization of sodium sulphate - a salt
g) NaCl -> Na+ + Cl- ionization of the salt, Sodium Chloride
Explanation:
Salts ionize at different rates; acids or bases dissociate; these are mostly strong acids and NaOH, a strong base.
Answer:
375 mL
Explanation:
M1*V1 = M2*V2
M1 = 1.00 M
V1 = ?
M2 = 0.750 M
V2 = 0.500 L
1.00 M * V1 = 0.750 M * 0.500 L
V1 = 0.750*0.500/1.00 = 0.375 L = 375 mL
Answer:
V = 4/3 * 3.1416 * (37x10-10)3
V = 2.12x10-25 cm3
d = m/V
d = 1.67x10-24 / 2.12x10-25 = 7.87 g/cm3
The difference in temperature, let's convert F to ºC:
ºC = -80-32/1.8 = -62.22 ºC
dT = -92.6 + 62.2 = -30.4 ºC
Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400
