Consider two solutions: solution x has a ph of 4; solution y has a ph of 7. from this information, we can reasonably conclude th
1 answer:
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a) To find the mass after t years:
we will use this formula:
A = Ao / 2^n
when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)
b) Mass after 90 y :
by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mg
C) Time for 1 mg remaining:
when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
we will use this formula:
A = Ao / 2^n
when A =the amount remaining
and Ao = the initial amount
and n = t / t(1/2)
by substitution:
∴ A = 200 mg/ 2^(t/30y)
b) Mass after 90 y :
by using the previous formula and substitute t by 90 y
A = 200mg/ 2^(90y/30y)
∴ A = 25 mg
C) Time for 1 mg remaining:
when A= Ao/ 2^(t/t(1/2)
so, by substitution:
1 mg = 200 mg / 2^(t/30y)
∴2^(t/30y) = 200 mg by solving for t
∴ t = 229 y
Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.