Answer:
We need 0.375 mol of CH3OH to prepare the solution
Explanation:
For the problem they give us the following data:
Solution concentration 0,75 M
Mass of Solvent is 0,5Kg
knowing that the density of water is 1g / mL, we find the volume of water:

Now, find moles of
are needed using the molarity equation:
therefore the solution is prepared using 0.5 L of H2O and 0.375 moles of CH3OH, resulting in a concentration of 0,75M
Answer:
A mole (mol) is the amount of a substance that contains 6.02 × 10 23 representative particles of that substance. The mole is the SI unit for the amount of a substance. There are, therefore, 6.02 × 10 23 water molecules in a mole of water molecules. Water (H2O) is made from 2 atoms of hydrogen and 1 atom of oxygen.
Answer : The concentration of A after 80 min is, 0.100 M
Explanation :
Half-life = 20 min
First we have to calculate the rate constant, we use the formula :



Expression for rate law for first order kinetics is given by:

where,
k = rate constant = 
t = time passed by the sample = 80 min
a = initial amount of the reactant = 1.6 M
a - x = amount left after decay process = ?
Now put all the given values in above equation, we get


Therefore, the concentration of A after 80 min is, 0.100 M
Answer:
a. 1 x 10^8
Explanation:
100 MHz = 100,000,000 Hz = 10^8 Hz
Oxidation number of fe in the compound...,
Let the oxidation number of (Fe) be x
The oxidation number of oxygen(o) is (-2)
; 3x + 4(-2) = 0...where zero is the total charge on the compound(fe3o4)
; 3x = 8...then divide both sides by 3
Oxidation number of fe is( 2.67 )