Interphase is not a phase of mitosis.
Answer:
for given question is 2.79 and 
is 0.52
{i- vant hoff’s constant ; Kb- constant ; m molarity }
M = no. of moles of the solute present in one kg of solution
Let the weight of amount of solute be “w” and its molecular mass be “M”
Let the mass of the solvent in the given question be “x”
Answer:
3.052 × 10^24 particles
Explanation:
To get the number of particles (nA) in a substance, we multiply the number of moles of the substance by Avogadro's number (6.02 × 10^23)
The mass of Li2O given in this question is as follows: 151grams.
To convert this mass value to moles, we use;
moles = mass/molar mass
Molar mass of Li2O = 6.9(2) + 16
= 13.8 + 16
= 29.8g/mol
Mole = 151/29.8g
mole = 5.07moles
number of particles (nA) of Li2O = 5.07 × 6.02 × 10^23
= 30.52 × 10^23
= 3.052 × 10^24 particles.
Answer:
Approximately 
.
Explanation:
Make use of the molar mass data (
) to calculate the number of moles of molecules in that 
of 
:
.
Make sure that the equation for this reaction is balanced.
Coefficient of in this equation: 
.
Coefficient of 
in this equation: .
In other words, for every two moles of that this reaction consumes, two moles of would be produced.
Equivalently, for every mole of that this reaction consumes, one mole of would be produced.
Hence the ratio: 
.
Apply this ratio to find the number of moles of that this reaction would have produced:
.
