Answer: 94.13 L
Explanation: In STP in an ideal gas there is a standard value for both temperature and pressure. At STP,pressure is equal to 1atm and the temperature at 0°C is equal to 273.15K. This problem is an ideal gas so we use PV=nRT where R is a constant R= 0.08205 L.atm/mol.K.
To find volume, derive the equation, it becomes V=nRT/P. Substitute the values. V= 4.20 mol( 0.08205L.atm/mol.K)(273.15K) / 1 atm = 94.13 L. The mole units, atm and K will be cancelled out and L will be the remaining unit which is for volume.
Answer:
The percent isotopic abundance of C- 12 is 98.93 %
The percent isotopic abundance of C- 13 is 1.07 %
Explanation:
we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)
First of all we will set the fraction for both isotopes
X for the isotopes having mass 13.003355
1-x for isotopes having mass 12
The average atomic mass of carbon is 12.0107
we will use the following equation,
13.003355x + 12 (1-x) = 12.0107
13.003355x + 12 - 12x = 12.0107
13.003355x- 12x = 12.0107 -12
1.003355x = 0.0107
x= 0.0107/1.003355
x= 0.0107
0.0107 × 100 = 1.07 %
1.07 % is abundance of C-13 because we solve the fraction x.
now we will calculate the abundance of C-12.
(1-x)
1-0.0107 =0.9893
0.9893 × 100= 98.93 %
98.93 % for C-12.
<span>Mass of the solution = 0.17m
Kb for C6H5NH2 = 3.8 x 10^-10
We know Ka for C6H5NH2 = 1.78x10^-11
We have Kw = Ka x Kb => Ka = Kw / Kb
=> (C2H5NH2)(H3O^+)/(C2H5NH3^+) => 1.78x10^-11 = K^2 / 0.17
K^2 = 3 x 10^-12 => K = 1.73 x 10^-6.
pH = -log(Kw(H3O^+)) = -log(1.73 x 10^-6) = 5.76</span>
Answer:
35.9%
Explanation:
The percent volume of the coffee solution can be calculated as follows:
% volume of coffee solution = volume of coffee/total volume of coffee solution × 100
According to this question, a cup of coffee has 71 mL of coffee and 127 mL of water. This means that, the total volume of coffee solution is;
71mL + 127mL = 198mL
% volume = 71/198 × 100
= 0.359 × 100
Percent volume of coffee solution = 35.9%
