5.512 litres is the volume of 15.2 grams of sulphur dioxide gas at STP.
Explanation:
Data given:
mass of sulphur dioxide = 15.2 grams
conditions is at STP whech means volume = 22.4 litres
atomic mass of sulphur dioxide = 64.06 grams/mole
Number of moles is calculated as:
number of moles = 
Putting the values in the equation:
number of moles = 
= 0.23 moles
Assuming that sulphur dioxide behaves as an ideal gas, we can calculate the volume as:
When 1 mole of sulphur dioxide occupies 22.4 litres at STP
Then 0.23 moles of sulphur dioxide occupies 22.4 x 0.23
= 5.152 litres is the volume.
Answer:
1.03
Explanation:
You would take 0.00103 and move the decimal like this; 0001.03, we wouldn't have the zeroes in front of the one, as it can throw us off. Therefore, your answer would be 1.03.
Hope I helped! Don't hesitate to let me know if I made a mistake.
The correct answer is 12.2% BaO.
The solution is found by dividing the mass of the BaO, which is 25.8 grams, by the total mass of the solution, which is 212 grams, then multiplying it by 100 to get the percentage:
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
