Answer:
The compound that will accumulate is B.
Explanation:
Enzymes in any biochemical process are the protein molecules that speed up the rate of the biological reaction. It lowers down the activation energy of the reactant molecules.
Since from the question:
<u>A + E1 –> B + E2 –> C</u>
Given that A, B and C are the molecules in the pathway and E1 and E2 are the enzymes.
<u>This means that A undergoes some biological reaction in which E1 is acting a catalyst to give B and which further undergoing another biological reaction to give C in the presence of enzyme E2.</u>
Since, E2 is inactive,
<u>A will form B but B will not be able to undergo further reaction and thus, C can not be formed.</u>
<u>Thus A is consumed and C is not formed. The compound that will accumulate is B.</u>
C6H15, C2H5 has a molar mass of 29g/mol. 87 divided by 29 is 3. Then multiply each element subscript by 3
Answer:
THE MOLAR MASS OF THE UNKNOWN MOLECULAR SUBSTANCE IS 200 G/MOL.
Explanation:
Mass of the unknown substance = 0.50 g
Freezing point of the solution = 3.9 °C
Freezing point of pure benzene = 5.5 °C
Freezing point dissociation constant Kf = 5.12°C/m
First, calculate the temperature difference between the freezing point of pure benzene and the final solution freezing point.
Change in temperature = 5.5 -3.9 = 1.6 °C
Next is to calculate the number of moles or molarity of the compound that dissolved.
Using the formula:
Δt = i Kf m
Assume i = 1
So,
1.6 °C = 1 * 5.12 * x/ 0.005 kg of benzene
x = 1.6 * 0.008 / 5.12
x = 0.0128 / 5.12
x = 0.0025 moles.
Next is to calculate the molar mass using the formula, molarity = mass / molar mass
Molar mass = mass / molarity
Molar mass = 0.50 g /0.0025
Molar mass = 200 g/mol
Hence, the molar mass of the unknown compound is 200 g/mol
Answer:
:)
Explanation:
The nucleus in biology is usually referring to that of
living cells found in organisms in which contains DNA whereas the nucleus in chemistry is usually referring to that of atoms in which contains subatomic particles such as the proton and neutron.