2K + 2H2O -----> 2KOH + H2
The question is telling you what are the reactants and products .
It is a solution, because it's the alloy of tin and copper
The radioactive decay follows first order kinetics
The integrated rate law expression for first order rate is
ln(A / At) = kt
Where A= initial concentration
At=concentration at time "t"
t = time
K= rate constant
ln(50 / 0.0488) = k X 38
Therefore
K = 0.1824 day-1
The relation between rate constant and half life is
So half life = 0.693 / 0.1824 = 3.8 days
So isotope must be Radon-222
Answer:
a) 0.525 mol
b) 0.525 mol
c) 0.236 mol
Explanation:
The combustion reactions (partial and total) will be:
C₇H₁₆ + (15/2)O₂ → 7CO + 8H₂O
C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O
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2C₇H₁₆ + (37/2)O₂ → 7CO + 7CO₂ + 16H₂O
It means that the reaction will form 50% of each gas.
a) 0.525 mol of CO
b) 0.525 mol of CO₂
c) The molar mass of heptane is: 7*12 g/mol of C + 16*1 g/mol of H = 100 g/mol
So, the number of moles is the mass divided by the molar mass:
n = 11.5/100 = 0.115 mol
For the stoichiometry:
2 mol of C₇H₁₆ -------------- (37/2) mol of O₂
0.115 mol of C₇H₁₆ --------- x
By a simple direct three rule:
2x = 2.1275
x = 1.064 mol of O₂
Which is the moles of oxygen that reacts, so are leftover:
1.3 - 1.064 = 0.236 mol of O₂
Answer:
Got your back
Explanation:
If the ions derived from different atoms are isoelectronic species, then they all have same number of electrons in their electronic shells and will have got same electronic configuration but their nuclear charge will differ because of their difference in number of protons in the nucleus. With increase in number of protons in the nucleus the electrons are more attracted towards nucleus thereby causing the decrease in ionic radius. On this principle our problem will be solved
The given ions are
7N-3
→no. of proton
=7
and
no of electron
=10
8O-2
→
no. of proton
=8
and
no of electron
=10
9F-→
no. of proton=9
and no of electron=10
11Na→
no. of proton=11
and no of electron=10
12 Mg-3→
no. of proton=12 and
no of electron=10
Hence the increasing order of ionic radius is
12Mg-3<11Na+<9F-<8O-2<7N-3
To rmember ->For isoelectronic species lower the nuclear charge higher the radius
