Answer:
10.6 g CO₂
Explanation:
You have not been given a limiting reagent. Therefore, to find the maximum amount of CO₂, you need to convert the masses of both reactants to CO₂. The smaller amount of CO₂ produced will be the accurate amount. This is because that amount is all the corresponding reactant can produce before it runs out.
To find the mass of CO₂, you need to (1) convert grams C₂H₂/O₂ to moles (via molar mass), then (2) convert moles C₂H₂/O₂ to moles CO₂ (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles CO₂ to grams (via molar mass). *I had to guess the chemical reaction because the reaction coefficients are necessary in calculating the mass of CO₂.*
C₂H₂ + O₂ ----> 2 CO₂ + H₂
9.31 g C₂H₂ 1 mole 2 moles CO₂ 44.0095 g
------------------ x ------------------- x ---------------------- x ------------------- =
26.0373 g 1 mole C₂H₂ 1 mole
= 31.5 g CO₂
3.8 g O₂ 1 mole 2 moles CO₂ 44.0095 g
------------- x -------------------- x ---------------------- x -------------------- =
31.9988 g 1 mole O₂ 1 mole
= 10.6 g CO₂
10.6 g CO₂ is the maximum amount of CO₂ that can be produced. In other words, the entire 3.8 g O₂ will be used up in the reaction before all of the 9.31 g C₂H₂ will be used.
Answer:
The United States customary system aka USCS or USC?
Answer:
a) 1512000 Joules
b) 5040 seconds = 84 minutes = 1.4 hours
Explanation:
Power saved y replacing bulbs = 60-18 = 42 W = 42 J/s
Time the bulb is used for = 10 hours
Energy saved during this time
42×10×60×60 = 1512000 Joules
Saved energy by replacing standard incandescent lightbulbs with energy-efficient compact fluorescent lightbulbs in 10 hours is 1512000 Joules
b) Power the plasma TV uses = 300 W = J/s
\frac{1512000}{300}=5040\ s3001512000=5040 s
Time a plasma TV can be used for with the saved energy is 5040 seconds = 84 minutes = 1.4 hours.
Had to look for the options and here is the answer. The concentration of glucose carriers in which the glucose concentration is being decreased to zero is 400. This level is already classified as above normal of the rate of glucose concentration. Hope this helps.
Answer: about 15 miles or exactly 15.53 miles
Explanation:
