Answer:
Small holes in plants that allow carbon dioxide in and oxygen and water vapor out
Explanation:
Stomata are tiny holes that open and close for the plant to breathe.
The output density is given as kg/m 3, lb/ft 3, lb/gal(US liq) and sl/ft 3. Specific weight is given as N/m 3 and lb f / ft 3.
Answer : The value of work done by an ideal gas is, 37.9 J
Explanation :
Formula used :
Expansion work = External pressure of gas × Volume of gas
Expansion work = 1.50 atm × 0.25 L
Expansion work = 0.375 L.atm
Conversion used : (1 L.atm = 101.3 J)
Expansion work = 0.375 × 101.3 = 37.9 J
Therefore, the value of work done by an ideal gas is, 37.9 J
Answer:
C2H4O2 + 2O2 ==> 2CO2 + 2H2O
Explanation:
Start with the carbons
C2H4O2 + O2 ==> 2CO2 + H2O There are 2 on the left, so you need 2 on the right.
Next deal with the hydrogens. You have 4 on the left so you have to make 4 on the right
C2H4O2 + O2 ==> 2CO2 + 2H2O
The oxygens are the real devil in this question. Be careful how you handle them. There are 2 * 2 = 4 from the CO2 and 2*1 = 2 from the water. The total is 6
Now you can't just put a 3 in front of the O2. There are 2 in the given chemical. So you don't need 6. You need 6 -2 = 4. But the oxygen is O2. You have to divide the 4 by 2 to get 2
C2H4O2 + 2O2 ==> 2CO2 + 2H2O
And that's your answer.
