Write an balance the equation
Na2O + H2O -> 2 NaOH
Calculate the molecular mass of Na2O and NaOH from the atomic mass from the periodic table.
Na = 23
O=16
H=1
Na2O = 23 * 2 + 16 = 62
NaOH = 23+16+1= 40
For the stoichiometry of the reaction one mole of Na2O = 62g produce two mol of NaOH = 2* 40= 80 g
120 g Na2O x 80g NaOH / 62g Na2O=
154.8 g NaOH
Answer: Heat of the solution = mass water × specific heat water × change in temperature
mass water = 260ml (1.00g/ml ) = 260g
specific heat of water = c(water) = 4.184J/ g°C
Heat change of water = final temperature - initial temperature
= 26.5 - 21.2
= 5.3 °C
H = 260 g ( 4.184J/g°C ) (5.3°C) = 5765J
Molar heat = 
= 16473J/mol
Explanation: finding molar heat requires first to look at specific heat of water and the change of water temperature
Answer:
— The molality of chloride ions in 300g of water is. A) 1.00 molal. B) 0.500 molal. C) 0.0553 molal. D) 0.111 molal.
Answer:
Case 1:
X = Any element from Group I
i) H
ii) Li
iii) Na
iv) K
v) Rb
vi) Cs
Y = 1
Case 2:
X = Any element from Group II
i) Be
ii) Mg
iii) Ca
iv) Sr
v) Ba
vi) Ra
Y = 2
Case 3:
X = Any element from Group III
i) B
ii) Al
iii) Ga
iv) In
v) Ti
Y = 3
Explanation:
The general formula given is as follow,
XCly
So, if X has +1 oxidation state, then it will require only one Cl atom with oxidation number -1 to form a neutral compound, therefore, y = 1.
If X has +2 oxidation state, then it will require two Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 2.
If X has +3 oxidation state, then it will require three Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 3.
Answer:
Q = 96.6 j
Explanation:
Given data:
Heat required = ?
Initial temperature = 19°C
Final temperature = 33°C
Mass of disc = 3.0 g
Specific heat capacity = 2.3 J/g.°C
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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 = 33°C - 19°C
ΔT = 14°C
Q = 3.0 g×2.3 J/g.°C × 14°C
Q = 96.6 j
