Na₂CO₃(s) → 2Na⁺(aq) + CO₃²⁻(aq)
The sodium carbonate formed from a strong base and a weak acid. Hydrolysis is subjected to the anion of a weak acid.
CO₃²⁻ + H₂O ⇄ HCO₃⁻ + OH⁻
HCO₃⁻ + H₂O ⇄ H₂CO₃ + OH⁻
pH>7 alkaline solution
2Na⁺ + CO₃²⁻ + 2H₂O ⇄ 2Na⁺ + 2OH⁻ + H₂CO₃
Answer:
b. lithium
Explanation:
furthest to the left in second period
Answer:
378mL
Explanation:
The following data were obtained from the question:
Pressure (P) = 99.19 kPa
Temperature (T) = 28°C
Number of mole (n) = 0.015 mole
Volume (V) =...?
Next, we shall convert the pressure and temperature to appropriate units. This is illustrated below:
For Pressure:
101.325 KPa = 1 atm
Therefore, 99.19 kPa = 99.19/101.325 = 0.98 atm
For Temperature:
T(K) = T(°C) + 273
T(°C) = 28°C
T(K) = 28°C + 273 = 301K.
Next we shall determine the volume of N2. The volume of N2 can be obtained by using the ideal gas equation as shown below:
PV = nRT
Pressure (P) = 0.98 atm
Temperature (T) = 301K
Number of mole (n) = 0.015 mole
Gas constant (R) = 0.0821atm.L/Kmol.
Volume (V) =...?
0.98 x V = 0.015 x 0.0821 x 301
Divide both side by 0.98
V = (0.015 x 0.0821 x 301) /0.98
V = 0.378 L
Finally, we shall convert 0.378 L to millilitres (mL). This is illustrated below:
1L = 1000mL
Therefore, 0.378L = 0.378 x 1000 = 378mL
Therefore, the volume of N2 collected is 378mL
Answer:
0.0738 M
Explanation:
HNO3 +LiOH = LiNO3 + H2O
Number of moles HNO3 = number of moles LiOH
M(HNO3)*V(HNO3) = M(LiOH)*M(LiOH)
M(HNO3)*50.0mL = 0.100M*36.90 mL
M(HNO3) = 0.100*36.90/50.0 M = 0.0738 M
