To determine the number of potassium laid side by side by a given distance, we simply divide the total distance to the diameter of each atom. The diameter is twice the radius of the atom. We calculate as follows:
number of atoms = 4770 / 231x10^-12 = 2.06x10^13 atoms
Answer:
(A) The work done by the system is -101.325J
(B) The workdone by the system is -90.75J
Explanation:
(A) Workdone = -PΔV
Given that A = 100cm2 = 0.01m2
distance d = 10cm = 0.1m
ΔV= Area × distance
ΔV= 0.01 ×0.1
ΔV = 0.001m3
P= external pressure = 1atm = 101325Pa
Workdone = -0.001 × 101325
W= - 101.325Pa m3
1Pam3 = 1J
Therefore W = - 101.325J
The work done on the system is -101.325J
(B) Workdone = -PΔV
Given that A = 50cm2 = 0.005m2
distance d = 15cm = 0.15m
ΔV= Area × distance
ΔV= 0.005×0.15
ΔV = 0.00075m3
P=121kPa = 121000Pa
W= - 121000 × 0.00075
W= -90.75Pa m3
1Pam3 = 1J
W = - 90.75J
The woekdone by the system is -90.75J
Answer:
Gamma
Explanation:
I'm not sure how to do it without calculations but:
E=hv
7*10^7 J/mol=6.626*10^34 Js * v
v=1*10^41
Gamma rays.
More here: https://www.hasd.org/faculty/AndrewSchweitzer/spectroscopy.pdf
Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
N₀ is the number of C-14 atoms per kg of carbon in the original sample at time = Os when its carbon was of the same kind as that present in the atmosphere today. After time ts, due to radioactive decay, the number of C-14 atoms per kg of carbon is the same sample which has decreased to N. λ is the radioactive decay constant.
Therefore N = N₀e-λt which is the radioactive decay equation,
N₀/N = eλt In (N₀.N= λt. This is the equation 1
The mass of carbon which is present in the sample os mc kg. So the sample has a radioactivity of A/mc decay is/kg. r is the mass of C-14 in original sample at t= 0 per total mass of carbon in a sample which is equal to [(total number of C-14 atoms in the sample at t m=m 0) × ma]/ total mass of carbon in the sample.
Now that the total number of C-14 atoms in the sample at t= 0/ total mass of carbon in sample = N₀ then r = N₀×ma
So N₀ = r/ma. this equation 2.
The activity of the radioactive substance is directly proportional to the number of atoms present at the time.
Activity = A number of decays/ sec = dN/dt = λ(number of atoms of C-14 present at time t) =
λ₁(N×mc). By rearranging we get N = A/(λmc) this is equation 3.
By plugging in equation 2 and 3 and solve t to get
t = 1/λ In (rλmc/m₀A).
