Why is a thermos considered a (mostly) closed system? Question 3 options: different types of liquids stay hot thermoses come in
1 answer:
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<u>Answer:</u> The chemical equations are written below.
<u>Explanation:</u>
<u>For a:</u> Methane reacts with oxygen gas to produce carbon dioxide and water.
Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water
The chemical equation for the combustion of methane follows:
- <u>For b:</u> Butane reacts with oxygen gas to produce carbon dioxide and water.
This is also an example of combustion reaction.
The chemical equation for the combustion of butane follows:
- <u>For c:</u> An aqueous solution of sulfuric acid reacts with aqueous potassium hydroxide to produce potassium sulfate and water.
When an acid reacts with a base, it leads to the formation of salt and water. This reaction is known as neutralization reaction
The chemical equation for the reaction of potassium hydroxide and sulfuric acid follows:
Hence, the chemical equations are written above.
Answer:
No. If you were to set up what the friend said it would look like this:
19.7g × ==>
×
The above does not look right because the grams do not cancel out. If it's not cancelled, it would be included in your final answer, but you're looking for moles not grams.
When setting up a stoichiometry equation, you have to put the same unit of measurement (ex- grams, cm, mm, etc.) on the numerator of one side and the denominator of the other side.
This would cancel out the unit.
For example: If you wanted to find out how much of 320 cm are in a meter.
*there are 100 cm in 1 meter*
(1. always start with the given number!)
(2. set up the next fraction where you can cancel out cm)
×
= 3.2 m
Going back, your friend would have to switch the units from grams/moles to moles/grams. It would cancel out grams, which would leave moles. Therefore, moles will be included in your final answer.
×
= _?_ moles
The Kb for ammonia is 1.8 X 10⁻⁵
The ammonia will dissociate as
NH₃ + H₂O ---> NH₄⁺ + OH⁻
initial 0.350 0 0
Change -x +x +x
Equilibrium 0.35-x x x
Putting values
We may ignore x in denominator as Kb is very low
therefore
x² = 1.8 X 10⁻⁵ X 0.35 = 0.63 X 10⁻⁵
x = 2.51 X 10⁻³ M
[OH⁻] = 2.51 X 10⁻³
pOH = -log[OH⁻] = 2.6
pH = 14- 2.6 = 11.4
b) after addition of 18 mL of HCl
The moles of acid added = molarity X volume = 0.5 x 18 = 9 mmol
the moles of base present = molarity X volume = 0.35 X 45 = 15.75 mmol
so moles of base left = 6.75 mmol
moles of salt formed = 9 mmol
This will results in formation of buffer
the pH of buffer is calculated using
pOH = pKb + log [salt] / [base]
pOH = 4.74 + log [9/6.75] = 4.86
pH = 9.14
c) on addition of 33 mL of HCl
The moles of acid added = molarity X volume = 0.5 x 33 = 16.5 mmol
the moles of base present = molarity X volume = 0.35 X 45 = 15.75 mmol
Hence the acid will completely neutralize the base
The acid left = 16.5 - 15.75 = 0.75 mmol
total volume = 33 + 45 = 75mL
[HCl] = 0.75 / 75 = 0.01 M
pH = -log[0.01] = 2