Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
4.8 g of Ca was added therefore mass present of Ca is 4.8 g
molar mass of Ca is 40 g/mol
molar mass is the mass of 1 mol of Ca
therefore if we substitute these values in the equation
number of moles of Ca = 4.8 g / 40 g/mol = 0.12 mol
0.12 mol of Ca is present
q2)
next we are asked to calculate the number of moles of water present
again we can use the same equation to find the number of moles of water
number of moles = mass present / molar mass
3.6 g of water is present
sum of the products of the molar masses of the individual elements by the number of atoms
H - 1 g/mol and O - 16 g/mol
molar mass of water = (1 g/mol x 2 ) + 16 g/mol = 18 g/mol
molar mass of H₂O is 18 g/mol
therefore number of moles of water = 3.6 g / 18 g/mol = 0.2 mol
0.2 mol of water is present
Answer:
The half-life varies depending on the isotope.
Half-lives range from fractions of a second to billions of years.
The half-life of a particular isotope is constant.
Answer:
Explanation:
In this reaction, we must exchange the amino group (
) for a fluorine atom (
). Also, the first step in this reaction is the addition of nitrous acid.
We must remember that the amino group in the presence of nitrous acid produces a diazonium salt. The 
group is a very good leaving group and many benzene derivatives can be produced from this intermediate (see figure 1).
If what we want is to bond a fluorine atom we must use to be able to produce m-ethylfluorobenzene (see figure 2).
I hope it helps!
Mass of methanol (CH3OH) = 1.922 g
Change in Temperature (t) = 4.20°C
Heat capacity of the bomb plus water = 10.4 KJ/oC
The heat absorbed by the bomb and water is equal to the product of the heat capacity and the temperature change.
Let’s assume that no heat is lost to the surroundings. First, let’s calculate the heat changes in the calorimeter. This is calculated using the formula shown below:
qcal = Ccalt
Where, qcal = heat of reaction
Ccal = heat capacity of calorimeter
t = change in temperature of the sample
Now, let’s calculate qcal:
qcal = (10.4 kJ/°C)(4.20°C)
= 43.68 kJ
Always qsys = qcal + qrxn = 0,
qrxn = -43.68 kJ
The heat change of the reaction is - 43.68 kJ which is the heat released by the combustion of 1.922 g of CH3OH. Therefore, the conversion factor is:
Answer:
The correct statements are given below
Explanation:
b Enoyl CoA isomerase an enzyme that converts cis double bonds to trans double bonds in fatty acid metabolism,bypasses a step that reduces Q,resulting in the higher ATP yield.
c Even chain fatty acids are oxidized to acetyl CoA in the beta oxidation pathway.
f The final round of beta oxidation foe a 13 carbon saturated fatty acid yields acetyl CoA and propionyl CoA a three carbon fragment.
