<span>Sodium carbonate (Na2CO3) reacts with acetic acid (CH3COOH) to form sodium acetate (NaCH3COO), carbon dioxide (CO2), and water (H2O). A chemist carries out this reaction in a bomb calorimeter. The reaction causes the temperature of a bomb calorimeter to decrease by 0.985 K. The calorimeter has a mass of 1.500 kg and a specific heat of 2.52 J/g K. What is the heat of reaction for this system? What equation should I use in this case? I've written down these notes: Steps: 1. Calculate the mass of the solution in total. 2. Convert mass to volume or vice versa if needed. 3. Calculate the temperature change of the solution. 4. Calculate the energy released by the reaction.</span>
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Answer:
1.552 moles
Explanation:
First, we'll begin by writing a balanced equation for the reaction showing how C8H18 is burn in air to produce CO2.
This is illustrated below:
2C8H18 + 25O2 -> 16CO2 + 18H2O
Next, let us calculate the number of mole of C8H18 present in 22.1g of C8H18. This is illustrated below:
Molar Mass of C8H18 = (12x8) + (18x1) = 96 + 18 = 114g/mol
Mass of C8H18 = 22.1g
Mole of C8H18 =..?
Number of mole = Mass /Molar Mass
Mole of C8H18 = 22.1/144
Mole of C8H18 = 0.194 mole
From the balanced equation above,
2 moles of C8H18 produced 16 moles of CO2.
Therefore, 0.194 mole of C8H18 will produce = (0.194x16)/2 = 1.552 moles of CO2.
Therefore, 1.552 moles of CO2 are emitted into the atmosphere when 22.1 g C8H18 is burned
Answer:
a) 21.133 minutes for series
b) 84.54 minutes when split into 4
Explanation:
Data provided:
Total flow, V₀ = 45 MGD
total volume of each basin, V = 2500 m³
Now,
1 MGD = 3785.4118 m³/day
also,
1 day = 1440 minutes
thus,
45 MGD = 45 × 3785.4118 m³/day
or
45 MGD = 170343.5305 m³/day
and,
170343.5305 m³/day in 'm³/min'
= 170343.5305 / 1440
= 118.2941 m³/min.
Therefore,
The retention time, t =
or
The retention time, t =
or
The retention time, t = 21.13 min
Hence,
for series of tanks the retention time is 21.133 min.
Now,
on splitting the tanks in 4
the volumetric flow rate will be
=
= 29.57 m³/min.
Therefore,
The retention time =
or
The retention time, t =
or
The retention time, t = 84.54 min