Answer: 8.24x10^5 years, or 8.24x10^3 centuries
Explanation: Sequential conversions to arrive at a total population rate of 2.31x10^10 atoms/sec for the entire population. Total seconds would be (6.02x10^23 atoms/2.31x10^10 atoms/sec) = 2.60x10^13 seconds. No breaks allowed, then (2.60x10^13 seconds)/(3.1536x10^7 sec/year)= 8.24x10^5 years.
Answer:
3.38%
Explanation:
Given that;
the mass of the whole milk sample = 100 g
volume of HCl = 100 mL = 0.1 L
molarity of HCl = 0.5 M
volume of NaOH = 34.50 mL = 0.0345 L
molarity of NaOH = 0.3512 M
Since we knew the molarity and volume of both HCl and NaOH; we can calculate their corresponding number of moles present.
So, number of moles of HCl = molarity of HCl × volume of HCl
number of moles of HCl = 0.5 M × 0.100 mL
= 0.05 mole
number of moles of NaOH = Molarity of NaOH × Volume of NaOH
number of moles of NaOH = 0.3512 M × 0.0345 L
= 0.012 mole
From the question, we can deduce that the number of HCl that is consumed by NH₃ is equal to the number of moles of HCl that is consumed by NaOH.
SO, number of moles of HCl consumed by NH₃ = Total moles of HCl - moles of HCl consumed by NaOH
= 0.05 mole - 0.012 mole
= 0.038 mole
However, to determine the mole of NH₃ present , we have:
number of moles of NH₃ present = number of moles of HCl consumed by NH₃ = 0.038
∴ the mass of Nitrogen with the molecular weight (14.0 g/mol) = 0.038 moles × 14.0 g/mol
= 0.530 g
Now, the percentage of Nitrogen can be calculated as;
%
the percentage of protein in the sample = CP × %age of N
where CP is given as 6.38
∴ the percentage of protein in the sample = 6.38 × 0.530%
the percentage of protein in the sample = 3.3814%
the percentage of protein in the sample = 3.38%
Answer:
a)
b)
c)
Explanation:
Hello,
In this case, in each reaction we must subtract the Gibbs free energy of formation the reactants to the Gibbs free energy of formation of the products considering each species stoichiometric coefficients. In such a way, the Gibbs free energy of formations are:
So we proceed as follows:
a)
b)
c)
Regards.
From the chemical formula the total mass of the compound can be determined. The mass of the 1 mole of the compound is its molar mass. The atom by which the molecule is generated, the mass of these atoms are expressed in terms of amu or atomic unit mass, but after formation of a molecule in a particular ratio the mass of each of the atom becomes the total molecular weight of the generated molecule. In this case the molecule posses three atoms X, Y and Z which are in a ratio of 2:2:7. Thus the chemical formula of the compound can be written as .
So the total mass of the compound in amu is {(2×47)+(2×42)+(7×16)} = {94+84+112}=290 amu.
Thus 1 mole of the compound contains 290 amu or 290 g by mass.
Henceforth 20 gram of the compound is equivalent to (20/290) = 0.068 mole.