Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Giv
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Answer:- Volume of the gas in the flask after the reaction is 156.0 L.
Solution:- The balanced equation for the combustion of ethane is:
From the balanced equation, ethane and oxygen react in 2:7 mol ratio or 2:7 volume ratio as we are assuming ideal behavior.
Let's see if any one of them is limiting by calculating the required volume of one for the other. Let's say we calculate required volume of oxygen for given 36.0 L of ethane as:
= 126 L
126 L of oxygen are required to react completely with 36.0 L of ethane but only 105.0 L of oxygen are available, It means oxygen is limiting reactant.
let's calculate the volumes of each product gas formed for 105.0 L of oxygen as:
= 60.0 L
Similarly, let's calculate the volume of water vapors formed:
= 90.0 L
Since ethane is present in excess, the remaining volume of it would also be present in the flask.
Let's first calculate how many liters of it were used to react with 105.0 L of oxygen and then subtract them from given volume of ethane to know it's remaining volume:
= 30.0 L
Excess volume of ethane = 36.0 L - 30.0 L = 6.0 L
Total volume of gas in the flask after reaction = 6.0 L + 60.0 L + 90.0 L = 156.0 L
Hence. the answer is 156.0 L.
Answer:
Across
2. Conduction.
3. Plates
4. Convection
5. Subduction
7. Earthquake
Down
1. Radioactive
6. Radiation
8. Sink
9. Slabpull
The clues are;
Across:
2. air molecules come in contact with warmer molecules
3. crust are made up of puzzle - like landmass called_____
4. rising and falling movement of material in the mantle
5. when tectonic plates push with each other
7. it is the result of movement of earth's plate
Down:
1. elements that play a vital role in Earth's internal heat
6. least important mode of heat transport
8. warm material rise; cool material______
9. heats build up underneath the crust
Answer:
NH₃/NH₄Cl
Explanation:
We can calculate the pH of a buffer using the Henderson-Hasselbalch's equation.
If the concentration of the acid is equal to that of the base, the pH will be equal to the pKa of the buffer. The optimum range of work of pH is pKa ± 1.
Let's consider the following buffers and their pKa.
- CH₃COONa/CH3COOH (pKa = 4.74)
- NH₃/NH₄Cl (pKa = 9.25)
- NaOCl/HOCl (pKa = 7.49)
- NaNO₂/HNO₂ (pKa = 3.35)
- NaCl/HCl Not a buffer
The optimum buffer is NH₃/NH₄Cl.