Answer:
Explanation:
Hello.
In this case, since the undergoing chemical reaction is only between the sodium bicarbonate and the acid HA:
For 0.561 g of yielded carbon dioxide (molar mass 44 g/mol), the following mass of sodium bicarbonate (molar mass 84 g/mol) that reacted was:
Considering the 1:1 mole ratio between CO2 and NaHCO3. Finally, the percent by mass of NaHCO3 is computed by dividing the mass of reacted NaHCO3 and t the mixture:
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Answer:
Part A: 36 MBq; Part B: 18 MBq
Explanation:
The half-life is the time it takes for half the substance to disappear.
The activity decreases by half every half-life
A =Ao(½)^n, where n is the number of half-lives.
Part A
3.0 da = 1 half-life
A = Ao(½) = ½ × 72 MBq = 36 MBq
Part B
6.0 da = 2 half-lives
A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq
It can grow, reproduce, and produce/use energy. Hope this helps!!!
2H₂ + O₂ = 2H₂O
n(H₂)=m(H₂)/M(H₂)
n(H₂)=5g/2.0g/mol=2.5 mol
n(O₂)=m(O₂)/M(O₂)
n(O₂)=40g/32.0g/mol=1.25 mol
H₂ : O₂ = 2 : 1
2.5 : 1.25 = 2 : 1
n(H₂O)=n(H₂)=2n(O₂)=2.5 mol
m(H₂O)=n(H₂O)M(H₂O)
m(H₂O)=2.5mol*18.0g/mol=45.0 g
It would be 23, s choice C.
