Answer:
In Fahrenheit it is 294.8°F and in Celsius it is 146°C.
Answer:
7.12 mm
Explanation:
From coulomb's law,
F = kqq'/r².................... Equation 1
Where F = force, k = proportionality constant, q and q' = The two point charges, r = distance between the two charges.
Make r the subject of the equation,
r = √(kqq'/F).......................... Equation 2
Given: q = q' = 75.0 nC = 75×10⁻⁹ C, F = 1.00 N
Constant: k = 9.0×10⁹ Nm²/C².
Substitute into equation 2
r = √[ (75×10⁻⁹ )²9.0×10⁹/1]
r = 75×10⁻⁹.√(9.0×10⁹)
r = (75×10⁻⁹)(9.49×10⁴)
r = 711.75×10⁻⁵
r = 7.12×10⁻³ m
r = 7.12 mm
Hence the distance between the point charge = 7.12 mm
Rubidium is an element that belongs to Group 1. As such it will have physical properties similar to the other Group 1 elements. Rubidium is below
Potassium in the periodic table but above
Cesium. As such it would be most like one of those two elements.
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
