Answer:
Original temperature (T1) = - 37.16°C
Explanation:
Given:
Gas pressure (P1) = 2.75 bar
Temperature (T2) = - 20°C
Gas pressure (P2) = 1.48 bar
Find:
Original temperature (T1)
Computation:
Using Gay-Lussac's Law
⇒ P1 / T1 = P2 / T2
⇒ 2.75 / T1 = 1.48 / (-20)
⇒ T1 = (2.75)(-20) / 1.48
⇒ T1 = -55 / 1.48
⇒ T1 = - 37.16°C
Original temperature (T1) = - 37.16°C
M1 = 17.45 M
M2 = 0.83 M
V2 = 250 ml
M1. V1= M2. V2
V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml
There is no path of electrons around the nucleus. There are however things called orbitals where you are likely to find electrons.
Gay Lussac's Law

Convert:
200°C = 200 + 273 = 473 K
50°C = 50 + 273 = 323 K
Input the value:
