Answer:
1.
molecules of CO₂
2. 10⁴ molecules of H₂O
3. 8.75×10³² molecules of C₆H₁₂O₆
Explanation:
1.
molecules of CO₂
2. 10⁴ molecules of H₂O
3. 8.75×10³² molecules of C₆H₁₂O₆
First, we have to get how many grams of C & H & O in the compound:
- the mass of C on CO2 = mass of CO2*molar mass of C /molar mass of CO2
= 0.5213 * 12 / 44 = 0.142 g
- the mass of H atom on H2O = mass of H2O*molar mass of H / molar mass of H2O
=0.2835 * 2 / 18 = 0.0315 g
- the mass of O = the total mass - the mass of C atom - the mass of H atom
= 0.3 - 0.142 - 0.0315 = 0.1265 g
Convert the mass to mole by divided by molar mass
C(0.142/12) H(0.0315/2) O(0.1265/16)
C(0.0118) H(0.01575) O(0.0079) by dividing by the smallest value 0.0079
C1.504 H3.99 O1 by rounding to the nearst fraction
C3/2 H4/1 )1/1 multiply by 2
∴ the emprical formula C3H8O2
B. They may be gases, liquids, or solids at room temp.
Answer:
2.12 moles of gas were added.
Explanation:
We can solve this problem by using<em> Avogadro's law</em>, which states that at constant temperature and pressure:
Where in this case:
We <u>input the data</u>:
- 6.13 L * n₂ = 11.3 L * 2.51 mol
As <em>4.63 moles is the final number of moles</em>, the number of moles added is: