Starfish, they break off a part of themselves and then it grows into another starfish, that’s how they reproduce asexually :)
Answer:
4 g OF IODINE-131 WILL REMAIN AFTER 32 DAYS.
Explanation:
Half life (t1/2) = 8 days
Original mass (No) = 64 g
Elapsed time (t) = 32 days
Mass remaining (Nt) = ?
Using the half life equation we can obtain the mass remaining (Nt)
Nt = No (1/2) ^t/t1/2
Substituting the values, we have;
Nt = 64 * ( 1/2 ) ^32/8
Nt = 64 * (1/2) ^4
Nt = 64 * 0.0625
Nt = 4 g
So therefore, 4 g of the iodine-131 sample will remain after 32 days with its half life of 8 days.
Answer:
3.0 moles Al₂O₃
Explanation:
We do not know which of the reactants is the limiting reactant. Therefore, you need to convert both of the given mole values into the product. This can be done using the mole-to-mole ratio made up of the balanced equation coefficients.
4 Al + 3 O₂ -----> 2 Al₂O₃
6.0 moles Al 2 moles Al₂O₃
---------------------- x ------------------------- = 3.0 moles Al₂O₃
4 moles Al
4.0 moles O₂ 2 moles Al₂O₃
---------------------- x ------------------------- = 2.7 moles Al₂O₃
3 moles O₂
As you can see, O₂ produces the smaller amount of product. This means O₂ is the limiting reactant. Remember, the limiting reactant is the reactant which runs out before the other reactant(s) are completely reacted. As such, the actual amount of Al₂O₃ produced is 2.7 moles.
However, since this problem is directly addressing how much Al₂O₃ is produced from Al, the answer you most likely are looking for is 3.0 moles Al₂O₃.
Answer:
a) rate law1 = k[NO2]²
b) rate law2 = k[NO][O3]
Explanation:
NO2(g) + CO(g) → NO(g) + CO2(g)
NO(g) + O3(g) → NO2(g) + O2(g)
When [NO2] in reaction 1 is doubled, the reaction quadruples
Rxn is second order.
rate law1= [NO2]^a [CO]^b
rate law1= [NO2]² [CO]^0
rate law1 = k[NO2]²
When [NO] in reaction 2 is doubled, the rate doubles.
Rxn is first order
The ratio is 1:1
this makes the rate law2 = k[NO][O3]
