In order to compute this, we must first take a couple of assumptions of:
1) The laboratory size so we can calculate its volume
2) The number of students working in the lab so we know the total gas produced
Let the lab be
11 m × 9 m × 6 m
The volume then computes to be:
594 m³
We know that
1 Liter is 1 dm³
1 m = 10 dm
1 m³ = 1000 dm³
Therefore, the room volume in liters is:
594,000 Liters
Let there be 30 students in the laboratory
Total gas being produced:
6 × 30
= 180 Liters
This works out to be:
0.03% of Hydrogen by volume
Therefore, there is no risk of explosion given our assumption of size and students.
Answer:
<h3>H2O (water)</h3><h3>N2 (nitrogen)</h3><h3>O3 (ozone)</h3><h3>CaO (calcium oxide)</h3><h3>
Explanation:</h3>
<u>H</u><u>OPE </u><u>IT</u><u> HELPS</u>
Answer:
4
Step-by-step explanation:
The reactions are:
Glycolysis: 1 glucose ⟶ 2 pyruvate
Link reaction: 2 × [1 pyruvate ⟶ 1 acetyl CoA]
Citric acid cycle: 2 × [1 AcetylCoA ⟶ 2 CO₂]
Now, add the reactions, cancelling species that occur on both sides of the reaction arrow,
1 glucose ⟶ <u>2 pyruvate
</u>
<u>2 pyruvate</u> ⟶ <u>2 acetyl CoA
</u>
<u>2 AcetylCoA </u>⟶ 4 CO₂
<em>Overall</em> : 1 glucose ⟶ 4 CO₂
For each mole of glucose, four molecules of CO₂ are released in the citric acid cycle.
Q=44.0 ct
p=3.51 g/cm³
Δq=0.200 g/ct
m=q×Δq
v=m/p=q×Δq/p
v=44.0×0.200/3.51≈2.5 cm³
C. 2.5 cm³
