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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Answer:
NaCl
Explanation:
Thus, for the compound between Na + and Cl −, we have the ionic formula NaCl (Figure 3.5 "NaCl = Table Salt").
Answer:
The first thing that you need to do here is to figure out the mass of the sample.
To do that, you can use its volume and the fact that aluminium is said to have a density of
2.702 g cm
−
3
, which implies that every
1 cm
3
of aluminium has a mass of
2.702 g
.
Explanation:
Answer:
Consequently, what happens when gas obtained by heating slaked lime and ammonium chloride is passed through copper sulphate solution? The HCl in the gas mixture will form hydrochloric and the H+ will react with some of the NH3(aq), forming NH4^+, and with some of the SO4^2-, forming HSO4^-.
