Answer:
74.12kg/day
Explanation:
Equation of reaction for the neutralization reaction of sodium hydroxide and sulfuric acid: 2NaOH + H2SO4 yields Na2SO4 + 2H2O
Mass of sulfuric acid produced per day = 90,800kg
Percentage of sulfuric acid in wastewater = 0.1%
Mass of sulfuric acid that ends up in wastewater per day = 0.1/100 × 90,800 = 90.8kg
From the equation of reaction, 2 moles of NaOH (80kg of NaOH) is required to neutralize 1 mole of H2SO4 (98kg of H2SO4)
80kg of NaOH is required to neutralize 98kg of sulfuric acid
90.8kg of sulfuric acid would be neutralized by (90.8×80)/98kg of NaOH = 74.14kg/day of NaOH
Answer:
The correct answer is 0.004382 kW/K, the second law is a=satisfied and established because it is a positive value.
Explanation:
Solution
From the question given we recall that,
The transferred heat rate is = 2kW
A reservoir cold at = 300K
The next step is to find the rate at which the entropy of the two reservoirs changes is kW/K
Given that:
Δs = Q/T This is the entropy formula,
Thus
Δs₁ = 2/ 300 = 0.006667 kW/K
Δs₂ = 2 / 875 =0.002285
Therefore,
Δs = 0.006667 - 0.002285
= 0.004382 kW/K
Yes, the second law is satisfied, because it is seen as positive.
Answer:
1 2 200 000 000 000 000 000 000 (20 zeros) molecules
Explanation:
Molar mass of 1-butanol = 74.121 g/mol
number of moles = mass given / molar mass
where mass given = 1.5 g
number of mole = 1.5 g / (74.121g/mol ) = 0.0202 moles
number of moles = number of molecules / Avogadro's constant
where Avogadro's = 6.022 × 10²³ mol⁻¹
number of moles = 6.022 × 10²³ × 0.0202 moles = 1 2 200 000 000 000 000 000 000 (20 zeros) molecules
