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: C) middle 50 percent of the data
The interquartile range (IQR) spans from the first quartile Q1 to the third quartile Q3.
25% of the data is below Q1 and 75% of the data is below Q3. The gap between the two endpoints consists of 75-25 = 50 percent of the data, or half of the data.
Answer:
1.2* 10³ rNe.
Explanation:
Given speed of neon=350 m/s
Un-certainity in speed= (0.01/100) *350 =0.035 m/s
As per heisenberg uncertainity principle
Δx*mΔv ≥\frac{h}{4\pi }
4π
h
..................(1)
mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg
6.22∗10
−23
20∗10
−3
=3.35∗10
−26
kg
substituating the values in eq. (1)
Δx =4.49*10^{-8}10
−8
m
In terms of rNe i.e 38 pm= 38*10^{-12}10
−12
Δx=\frac{4.49*10^{-8} }{38*10^{-12} }
38∗10
−12
4.49∗10
−8
=0.118*10^{4}10
4
* (rNe)
=1.18*10³ rN
= 1.2* 10³ rNe.
Explanation:
This is the answer
