Answer:
In oxidation reduction reactions, one species gets reduced by taking on electron(s) and another species gets oxidized by losing electrons. They also flow by a wire
Explanation:
To estimate the molar mass of the gas, we use Graham's law of effusion. This relates the rates of effusion of gases with their molar mass. We calculate as follows:
r1/r2 = √(m2/m1)
where r1 would be the effusion rate of the gas and r2 is for CO2, M1 is the molar mass of the gas and M2 would be the molar mass of CO2 (44.01 g/mol)
r1 = 1.6r2
1.6 = √(44.01 / m1)
m1 = 17.19 g/mol
16.94 grams-------------------
The atom consists of a nucleus with electrons orbiting at specific energy levels
4,000,000 L
1g/.001kg = x/8,000
= 8,000,000/2 = 4,000,000
