Answer:
1/32 of the original sample
Explanation:
We have to use the formula
N/No = (1/2)^t/t1/2
N= amount of radioactive sample left after n number of half lives
No= original amount of radioactive sample present
t= time taken for the amount of radioactive samples to reduce to N
t1/2= half-life of the radioactive sample
We have been told that t= five half lives. This implies that t= 5(t1/2)
N/No = (1/2)^5(t1/2)/t1/2
Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;
N/No= (1/2)^5
N/No = 1/32
Hence the fraction left is 1/32 of the original sample.
Answer:
T₂ = 182 K
Explanation:
Given that,
Initial pressure, P₁ = 120 kPa
Initial temperature, T₁ = 0˚C = 273 K
We need to find the final temperature when the pressure is 80 kPa.
We know that, Gay Lussac's Formula is :
So, the new temperature is equal to 182 K.
Mass of methane takne = 1.5g
moles of methane used = masss / molar mass = 1.5 / 16 = 0.094 moles
mass of water = 1000 g
Initial temperature of water = 25 C
final temperature = 37 C
specific heat of water = 4.184 J /g C
1) Heat absorbed by water = q =m• C• ΔT = 1000 X 4.184 x (37-25) = 50208 Joules
2) Heat absorbed by calorimeter = Heat capacity X ΔT = 695 X (37-25) = 8340 J
3) Total heat of combustion = heat absorbed by water + calorimeter = 50208 + 8340 = 58548 Joules
This heat is released by 0.094 moles of methane
So heat released by one mole of methane =
- 622851.06 Joules = 622.85 kJ / mole
4) standard enthalpy of combustion = -882 kJ / mole
Error = (882-622.85) X 100 / 882 = 24.84 %
Answer:
I believe it is Spruce, because of the shape of the leaves
