4 moles of hydrogen = 4 * 1.008 = 4.032 grams
1 mole of helium = 4.003 grams
Difference is 4.032 - 4.003
= 0.029 g
Answer: 406 hours
Explanation:

where Q= quantity of electricity in coloumbs
I = current in amperes = 39.5 A
t= time in seconds = ?
The deposition of copper at cathode is represented by:

Coloumb of electricity deposits 1 mole of copper
i.e. 63.5 g of copper is deposited by = 193000 Coloumb
Thus 19.0 kg or 19000 g of copper is deposited by =
Coloumb

(1hour=3600s)
Thus it will take 406 hours to plate 19.0 kg of copper onto the cathode if the current passed through the cell is held constant at 39.5 A
Here is the full question:
Air containing 0.04% carbon dioxide is pumped into a room whose volume is 6000 ft3. The air is pumped in at a rate of 2000 ft3/min, and the circulated air is then pumped out at the same rate. If there is an initial concentration of 0.2% carbon dioxide, determine the subsequent amount in the room at any time.
What is the concentration at 10 minutes? (Round your answer to three decimal places.
Answer:
0.046 %
Explanation:
The rate-in;

= 0.8
The rate-out
= 
= 
We can say that:

where;
A(0)= 0.2% × 6000
A(0)= 0.002 × 6000
A(0)= 12

Integration of the above linear equation =

so we have:



∴ 
Since A(0) = 12
Then;



Hence;



∴ the concentration at 10 minutes is ;
=
%
= 0.0456667 %
= 0.046% to three decimal places
Hi,
Two sodium atoms are needed to create an ionic bond with sulfur.