Answer:
d. Because those chemicals are easily made when CO2 reacts with water, forming H2CO3 (via carbonic anhydrase
32.8 g of Butane is required and 99.3 g of CO₂ is produced
<u>Explanation:</u>
The above mentioned reaction can be written as,
C₄H₁₀(g) + 13 O₂(g) → 4CO₂(g) + 5 H₂O(g) where ΔH (rxn)= -2658 kJ
It is given that 1.5 × 10³ kJ of energy is produced, the original reaction says that 2658 kJ of heat is produced, which means that less than one mole of butane is used in the reaction.
That is
of butane reacted
Now this moles is converted into mass by multiplying it with its molar mass = 0.564 mol × 58.122 g / mol
= 32.8 g of butane.
Mass of CO₂ produced = 0.564 ×44.01 g /mol × 4 mol
= 99.3 g of CO₂
Thus 32.8 g of Butane is required and 99.3 g of CO₂ is produced
Work = force*distance
Work = 670 * 5
Work = 3350 Nm
Answer: 1.027 x 10^6 g= 1027kg
In this question, you are given the volume of the blimp (2.027×10^5 ft^3) and the density of the gas(0.179g/L). To answer this question, you need to convert the volume unit into liter. The calculation would be: 2.027×10^5 ft^3 x 28.3168L/ft3= 57.398 x 10^5L= 5.74x10^6L
Then to find the mass, multiply the volume with the density. The calculation would be: 5.74x10^6L x 0.179g/L= 1.027 x 10^6 g= 1027kg
Answer: 6.26atm
Explanation:Please see attachment for explanation