Net overall dissociation:
Na2CO3 ---> 2Na(-) + CO3(2-)
*The ion charge is in parenthesis
Answer:
16 °C
Explanation:
Step 1: Given data
- Provided heat (Q): 811.68 J
- Mass of the metal (m): 95 g
- Specific heat capacity of the metal (c): 0.534 J/g.°C
Step 2: Calculate the temperature change (ΔT) experienced by the metal
We will use the following expression.
Q = c × m × ΔT
ΔT = Q/c × m
ΔT = 811.68 J/(0.534 J/g.°C) × 95 g = 16 °C
Answer:
35Cl = 75.9 %
37Cl = 24.1 %
Explanation:
Step 1: Data given
The relative atomic mass of Chlorine = 35.45 amu
Mass of the isotopes:
35Cl = 34.96885269 amu
37Cl = 36.96590258 amu
Step 2: Calculate percentage abundance
35.45 = x*34.96885269 + y*36.96590258
x+y = 1 x = 1-y
35.45 = (1-y)*34.96885269 + y*36.96590258
35.45 = 34.96885269 - 34.96885269y +36.96590258y
0.48114731 = 1,99704989y
y = 0.241 = 24.1 %
35Cl = 34.96885269 amu = 75.9 %
37Cl = 36.96590258 amu = 24.1 %
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