Answer:
V₂ → 106.6 mL
Explanation:
We apply the Ideal Gases Law to solve the problem. For the two situations:
P . V = n . R . T
Moles are still the same so → P. V / R. T = n
As R is a constant, the formula to solve this is: P . V / T
P₁ . V₁ / T₁ = P₂ .V₂ / T₂ Let's replace data:
(1.20 atm . 73mL) / 112°C = (0.55 atm . V₂) / 75°C
((87.6 mL.atm) / 112°C) . 75°C = 0.55 atm . V₂
58.66 mL.atm = 0.55 atm . V₂
58.66 mL.atm / 0.55 atm = V₂ → 106.6 mL
Answer:
1) 1,... 2
2) 18
3) n= 3 and I=1
Explanation:
1) when l= 0, its an s-sub-level, and only 1 orbital is possible which can carry only 2-electrons
2) the maximum number of electron is given by 2n^2= 2×3^2= 18
3) in 3p, the coefficient of p is the value of n= 3 and l-value of P is 1
Which type of radioactive decay produces particles with the most mass? Alpha.
Answer:
152 pm
Explanation:
According to the question, we can estimate the bond length from the given values of the atomic radii. This now is the upper limit of the bond length for the molecule.
Since we have that;
Atomic radius of H= 37.0 pm
Atomic radius of Br = 115.0 pm
Bond length = Atomic radius of H + Atomic radius of Br
Bond length = 37.0 pm + 115.0 pm
Bond length = 152 pm
Oxygen because it is on the left of the periodic table so it has a strong pull.
