H3O+(aq) + OH-(aq) --> 2H2O (l)
NaHCO3(s) --> NaH 2+ (aq) + CO3 2- (aq)
NaH 2+ (aq) + H2O (l) --> Na+ (aq) + H3O+ (aq)
H2O (l) + CO3 2- (aq) --> OH- (aq) + HCO3- (aq)
(I'm not completely sure if I did the third question right) I'm sorry if I got it wrong
Answer:
-26.125 kj
Explanation:
Given data:
Mass of water = 250.0 g
Initial temperature = 30.0°C
Final temperature = 5.0°C
Amount of energy lost = ?
Solution:
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = T2 - T1
ΔT = 5.0°C - 30.0°C
ΔT = -25°C
Specific heat of water is 4.18 j/g.°C
Now we will put the values in formula.
Q = m.c. ΔT
Q = 250.0 g × 4.18 j/g.°C × -25°C
Q = -26125 j
J to kJ
-26125 j ×1 kj /1000 j
-26.125 kj
they will both do the exact same thing, as long as they are bothh identical
Answer:
V = 134.5 L
Explanation:
Given data:
Number of moles of KClO₃ = 4 mol
Litters of oxygen produced at STP = ?
Solution:
Chemical equation:
2KClO₃ → 2KCl + 3O₂
Now we will compare the moles of KClO₃ with oxygen.
KClO₃ : O₂
2 : 3
4 ; 3/2×4 = 6 mol
Litters of oxygen at STP:
PV = nRT
V = nRT/P
V = 6 mol × 0.0821 atm.L/mol.K × 273 K / 1atm
V = 134.5 L / 1
V = 134.5 L
