Answer:
V₂ = 1.86 L
Explanation:
Given data:
Initial volume = 4.30 L
Initial pressure = 1 atm
Initial temperature = 273.15 K
Final temperature = 302 K
Final volume = ?
Final pressure = 2.56 atm
Solution:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
V₂ = P₁V₁T₂
/T₁ P₂
V₂ = 1 atm ×4.30 L × 302 K / 273.15 K × 2.56 atm
V₂ = 1298.6 atm.L.K / 699.26 K.atm
V₂ = 1.86 L
Explanation:
so for this u have to use this equation where
Moles = number of particle/6.02×10^23
= 3.045 × 10^24/6.02×10^23
= 5.0581
write it to 3 S.F so 5.06 moles
Answer:
0.453 moles
Explanation:
The balanced equation for the reaction is:
2Fe(s) + 3O2(g) ==> 2Fe2O3
From the equation, mass of O2 involved = 16 x 2 x 3 = 96g
mass of Fe2O3 involved = [(2x26) + 3 x 16] x 2
= 100g
Therefore 96g of O2 produced 100g of Fe2O3
32.2g of O2 Will produce 100x32.2/96
= 33.54g of Fe2O3
Converting it to mole using number of mole = mass/molar mass
but molar mass of Fe2O3 = 26 + (16 X 3)
= 74g/mole
Therefore number of mole of 33.54g of Fe2O3 = 33.54/74
= 0.453 moles
Answer:
320 g
Step-by-step explanation:
The half-life of Co-63 (5.3 yr) is the time it takes for half of it to decay.
After one half-life, half (50 %) of the original amount will remain.
After a second half-life, half of that amount (25 %) will remain, and so on.
We can construct a table as follows:
No. of Fraction Mass
half-lives t/yr Remaining Remaining/g
0 0 1
1 5.3 ½
2 10.6 ¼
3 15.9 ⅛ 40.0
4 21.2 ¹/₁₆
We see that 40.0 g remain after three half-lives.
This is one-eighth of the original mass.
The mass of the original sample was 8 × 40 g = 320 g
The frequency of a photon of red light with wavelength 4.50 x 10−7m is 6.67 x 1014Hz
