Answer:
a) if the liquid is not vaporized completely, then the condensed vapor in the flask contains the air which is initially occupied before the liquid is heated. When calculating the molar mass of the vapor the moles of air which are initially present are not excluded, so that the molar mass of the vapor would be an increase in value.
b) While weighing the condensed vapor, the flask should be dried. If the weighing flask is not dried then the water which is layered on the surface of the flask is also added to the mass of the vapor. Therefore, the mass of the vapor that is calculated would be increase.
c) When condensing the vapor, the stopper should not be removed from the flask, because the vapor will escape from the flask and a small amount of vapor will condense in the flask. Therefore, the mass of the condensed vapor would be In small value.
d) If all the liquid is vaporized, when the flask is removed before the vapor had reached the temperature of boiling water, then the boiling
temperature of that liquid would be lower than that of the boiling temperature of the water.Therefore, the liquid may have more volatility.
Answer:- 64015 J
Solution: There is 4250 mL of water in the calorimeter at 22.55 degree C.
density of water is 1 g per mL.
So, the mass of water = 
= 4250 g
Final temperature of water after adding the hot copper bar to it is 26.15 degree C.
So, 
for water = 26.15 - 22.55 = 3.60 degree C
Specific heat for water is 4.184 
The heat gained by water is calculated by using the formula:
where, q is the heat energy, m is mass and c is specific heat.
Let's plug in the values in the formula and do the calculations:
q = 64015 J
So, 64015 J of heat is gained by the water.
Answer:
C 8.09 SO2 gas
Explanation:
As we have the volume (3dm³ = 3L), temperature (25°C + 273 = 298K), and pressure (1atm), we can solve to moles of gas using:
PV = nRT
PV / RT = n
1atm*3L / 0.082atmL/molK*298K =¨
0.123 moles of gas you have.
Now, to convert these moles to mass we use molar mass (32g/mol for O2, 28g/mol for N2, 64g/mol for SO2, and 44g/mol for CO2).
Mass of 0.123 moles of these gases is:
O2 = 0.123 moles * 32g/mol = 3.94g of O2. A is wrong
N2 = 0.123 moles * 28g/mol = 3.4g of N2. B is wrong
SO2 = 0.123 moles * 64.1g/mol = 7.9g of SO2≈ 8.09g of SO2, C is possible
CO2 = 0.123 moles * 44g/mol = 5.4g of CO2. D is wrong
Right answer is:
<h3>C 8.09 SO2 gas
</h3>
3 Mg + 1 Fe2O3 →2 Fe + 3MgO
Type of Reaction: Single displacement.
