T(1/2)=1/(k[NO2])
[NO2]=1/(kt(1/2))
[NO2]=1/(0.54M-1 s-1*22s) seconds cancel
[NO2]=1/11.88=0.0842
inital mole= 0.0842*volume(2.5)=0.210 mols
Answer:
The answer to your question is: first two.
Explanation:
Codons that code for Arginine AGA and AGG
Then,
AGA AGG CGA CGC CGG CGU
The first two codons will code for Arginine
Answer:
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
Explanation:
Step 1: Data given
tetraphosphorus decoxide = P4O10
Molar mass of P4O10 = 283.89 g/mol
Mass of P4O10 = 4.5 grams
Number of Avogadro = 6.022 * 10^23 / mol
Step 2: Calculate moles of P4O10
Moles P4O10 = mass P4O10 / molar mass P4O10
Moles P4O10 = 4.5 grams / 283.89 g/mol
Moles = 0.016 moles
Step 3: Calculate moles of P
For 1 mol P4O10 we have 4 moles of phosphorus
For 0.016 moles P4O10 we have 4*0.016 = 0.064 moles P
Step 4: Calculate number of P atoms
Number of P atoms = moles P * number of Avogadro
Number of P atoms = 0.064 moles * 6.022*10^23
Number of P atoms = 3.85 * 10^22 atoms
In 4.5 grams of tetraphosphorus decoxide we have 3.85 * 10^22 phosphorus atoms
2AgNO3 + Ni2+ = Ni(NO3)2 + 2Ag<span>+</span>
From the reaction,
it can be seen that AgNO3 and Ni2+ has following amount of substance
relationshep:
n(AgNO3):n(Ni)=2:1
From the relationshep we can determinate requred moles of Ni2+:
n(AgNO3)=m/M= 15.5/169.87=0.09 moles
So, n (Ni)=n(AgNO3)/2=0.045 moles
Finaly needed mass of Ni2+ is:
m(Ni2+)=nxM=0,045x58.7=2.64g
