Answer:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is .
Explanation:
The combined gas equation is,
where,
= initial pressure of gas = 104 kPa
= final pressure of gas = 52 kPa
= initial volume of gas =
= final volume of gas = ?
= initial temperature of gas =
= final temperature of gas =
Now put all the given values in the above equation, we get:
The volume of air at where the pressure and temperature are 52 kPa, -5.0 ºC is .
<u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>
To find the mass of glucose produced, first you must know the balanced reaction. For this, the Law of Conservation of Matter is followed.
The law of conservation of matter states that since no atom can be created or destroyed in a chemical reaction, the number of atoms that are present in the reagents has to be equal to the number of atoms present in the products.
So, in this case, the balanced reaction is:
6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
By stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the amounts of moles of each reactant and product participate in the reaction:
So, you know that 2.20 g of CO₂ react, whose molar weight is 44.01 g/mole. By definition of molar mass, 1 mole of CO₂ has 44.01 g. So, the number of moles that 2.20 grams of the compound represent is calculated as:
moles of CO₂= 0.05 moles
Now you must follow the following rule of three: if by stoichiometry of the reaction 6 moles of CO₂ produce 1 mole of C₆H₁₂O₆, 0.05 moles of CO₂ produce how many moles of C₆H₁₂O₆?
moles of C₆H₁₂O₆= 8.33*10⁻³
Being the molar mass of glucose 180.18 g/mole, the mass that 8.33*10⁻³ moles of the compound represent is calculated as:
<em>mass of glucose= 1.5 grams</em>
Then, <u><em>1.5 grams of glucose is produced from 2.20 g of CO₂.</em></u>