It takes 21.3 days
<h3>Further explanation</h3>
Given
5 hr = 8 kg Alcohol
Required
Days to consume 1000 kg of glucose
Solution
Alcoholic fermentation⇒ glucose produces ethanol and carbon dioxide,
C₆H₁₂O₆ → 2 C₂H₅OH + 2CO₂
mol ethanol :
moles of glucose to produce 108.7 moles ethanol :
54.35 moles = 5 hours
moles of 1000 kg of glucose :
So for 5555.5 moles, it takes :
<h3>
Answer:</h3>
1.25 moles (R.T.P.) or 1.34 moles (S.T.P.)
<h3>
Explanation:</h3>
- 1 mole of a gas occupies a volume of 24 liters at room temperature and pressure (R.T.P.)
- On the other hand, 1 mole of a gas will occupy 22.4 Liters at standard temperature and pressure (S.T.P.)
Therefore, at R.T.P.
30.0 Liters will be equivalent to;
= 30.0 L ÷ 24 L
= 1.25 moles
At S.T.P
30.0 Liters will be equivalent to;
= 30.0 L ÷ 22.4 L
= 1.34 moles
Thus, 30.0 L of helium gas are equivalent to 1.25 moles of He at R.T.P. and 1.34 moles at S.T.P.
Answer: D i think
Explanation: Im sorry if it is wrong lol!! I pretty sure thats the answer...
True is does make up 90 percent
