5.6L of O2 means we have 0.25 moles of O2.
As, 1 mole has 6.023*10^23 molecules,
0.25 moles of O2 will have 0.25*6.023*10^23 molecules=1.50575*10^23 molecules
and as 1 molecule of O2 has 2 atoms, so, 1.50575*10^23 molecules will have 2*1.50575*10^23 atoms=3.0115*10^23 atoms of O.
Answer:
hinndndnnddnndndndnd do djfj
Explanation:
hdhdjdbdndndndjjddjdndjdjdndnndndndnd be rnnrbr
<h3>
Answer:</h3>
42960 years
<h3>
Explanation:</h3>
<u>We are given;</u>
- Remaining mass of C-14 in a bone is 0.3125 g
- Original mass of C-14 on the bone is 80.0 g
- Half life of C-14 is 5370 years
We are required to determine the age of the bone;
- Remaining mass = Original mass × 0.5^n , where n is the number of half lives.
Therefore;
0.3125 g = 80.0 g × 0.5^n
3.90625 × 10^-3 = 0.5^n
- Introducing logarithm on both sides;
log 3.90625 × 10^-3 = n log 0.5
Solving for n
n = log 3.90625 × 10^-3 ÷ log 0.5
= 8
- Therefore, the number of half lives is 8
- But, 1 half life is 5370 years
- Therefore;
Age of the rock = 5370 years × 8
= 42960 years
Thus, the bone is 42960 years old
Answer:
Ag+(aq) + Cl-(aq) —> AgCl(s)
Explanation:
2AgNO3(aq) + CaCl2(aq) —>2AgCl(s) + Ca(NO3)2(aq)
The balanced net ionic equation for the reaction above can be obtained as follow:
AgNO3(aq) and CaCl2(aq) will dissociate in solution as follow:
AgNO3(aq) —> Ag+(aq) + NO3-(aq)
CaCl2(aq) —> Ca2+(aq) + 2Cl-(aq)
AgNO3(aq) + CaCl2(aq) –>
2Ag+(aq) + 2NO3-(aq) + Ca2+(aq) + 2Cl-(aq) —> 2AgCl(s) + Ca2+(aq) + 2NO3-(aq)
Cancel out the spectator ions i.e Ca2+(aq) and 2NO3- to obtain the net ionic equation.
2Ag+(aq) + 2Cl-(aq) —> 2AgCl(s)
Divide through by 2
Ag+(aq) + Cl-(aq) —> AgCl(s)
The, the net ionic equation is
Ag+(aq) + Cl-(aq) —> AgCl(s)
