Answer: The volume in the balloon at the higher altitude is 260 L
Explanation:
The combined gas equation is,
where,
= initial pressure of gas = 760 torr
= final pressure of gas = 511 torr
= initial volume of gas = 233 L
= 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 in the balloon at the higher altitude is 260 L
I’m pretty sure it’s -2 my friend said it ♂️
Answer:
24.0 g C₃H₈
Explanation:
To find the mass of C₃H₈, you need to (1) convert grams CO/H₂ to moles CO/H₂ (via molar mass), then (2) convert moles CO/H₂ to moles C₃H₈ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles C₃H₈ to grams C₃H₈ (via molar mass). It is important to arrange the conversions/ratios in a way that allows for the cancellation of units (the desired unit should be in the numerator). The final answer should have 3 sig figs to reflect the sig figs in the given values.
Molar Mass (CO): 12.011 g/mol + 15.998 g/mol
Molar Mass (CO): 28.009 g/mol
Molar Mass (H₂): 2(1.008 g/mol)
Molar Mass (H₂): 2.016 g/mol
Molar Mass (C₃H₈): 3(12.011 g/mol) + 8(1.008 g/mol)
Molar Mass (C₃H₈): 44.097 g/mol
3 CO + 7 H₂ ----> 1 C₃H₈ + 3 H₂O
^ ^ ^
45.8 g CO 1 mole 1 mole C₃H₈ 44.097 g
----------------- x ------------------ x -------------------- x ------------------ =
28.009 g 3 moles CO 1 mole
= 24.0 g C₃H₈
87.3 g H₂ 1 mole 1 mole C₃H₈ 44.097 g
---------------- x --------------- x --------------------- x ----------------- =
2.016 g 7 moles H₂ 1 mole
= 273 g C₃H₈
It was necessary to find the mass of the products from both of the reactants because you did not know which one was the limiting reagent. The limiting reagent is the reactant which is completely used up first. Because CO produced the smaller amount of product, it must be the limiting reagent. Therefore, the actual amount of C₃H₈ produced is 24.0 grams.
Answer:
put the solution over ice.
