Answer:
NaS2 it is the most stable
Explanation:im not really sure about this one but i think it is either c or b
ANSWER:C im pretty sure.sorry if wrong
The reaction between NaOH and Cu(NO₃)₂ is as follows
2NaOH + Cu(NO₃)₂ ---> 2NaNO₃ + Cu(OH)₂
Q1)
stoichiometry of NaOH to Cu(NO₃)₂ is 2:1
this means that 2 mol of NaOH reacts with 1 mol of Cu(NO₃)₂
the mass of Cu(NO₃)₂ reacted - 0.8024 g
molar mass of Cu(NO₃)₂ is 187.56 g/mol
therefore the number of Cu(NO₃)₂ moles that have reacted
- 0.8024 g/ 187.56 g/mol = 0.00427 mol
according to the stoichiometry , number of NaOH moles - 0.00427 mol x 2
then number of NaOH moles that have reacted - 0.00855 mol
In a 3.0 M NaOH solution, 3 moles are in 1000 mL of solution
Then volume required for 0.00855 mol - 1000 x 0.00855 /3 = 2.85 mL
2.85 mL of 3.0 M NaOH is required for this reaction
Q2)
Assuming that there's 100 % yield of Cu(OH)₂ , we can directly calculate the mass of Cu(OH)₂ formed from the number of moles of reactants that were used up.
Stoichiometry of Cu(NO₃)₂ to Cu(OH)₂ is 1:1
this means that 1 mol of Cu(NO₃)₂ gives a yield of 1 mol of Cu(OH)₂
the number of Cu(NO₃)₂ moles that reacted - 0.00427 mol
Therefore an equal amount of moles of Cu(OH)₂ were formed
Then amount of Cu(OH)₂ moles produced - 0.00427 mol
Mass of Cu(OH)₂ formed - 0.00427 mol x 97.56 g/mol = 0.42 g
A mass of 0.42 g of Cu(OH)₂ was formed in this reaction
<span>2HBr→<span>H2</span>+B<span>r2 i believe </span></span>
Corrected question:
Bromine-88 is radioactive and has a half life of 16.3seconds. What percentage of a sample would be left after 35seconds? Round your answer to 2 significant digits.
Answer:
23% (3 significant digits)
Explanation:
where is mass remaining
is original mass
t1/2= 16.3s , t=35s
n =35/16.3 =2.147
=
0.2258*100% =22.5%
≈23%