Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
The given statement is True.
This is single replacement reaction.
As reactivity decreases down the group,
Iodine is less reactive than Chlorine.
And so Iodine cannot replace chlorine from FeCl2.
I2 + FeCl2 ---> no reaction (Iodine is less reactive than chlorine)
Answer:
The metric system uses units such as meter, liter, and gram to measure length, liquid volume, and mass, just as the U.S. customary system uses feet, quarts, and ounces to measure these.
Volume: 1 liter is a little more than 1 quart
Mass: 1 kilogram is a little more than 2 pounds
Length: 1 centimeter is a little less than half an
Explanation:
Answer : The molecular weight of a gas is, 128.9 g/mole
Explanation : Given,
Density of a gas = 5.75 g/L
First we have to calculate the moles of gas.
At STP,
As, 22.4 liter volume of gas present in 1 mole of gas
So, 1 liter volume of gas present in 
mole of gas
Now we have to calculate the molecular weight of a gas.
Formula used :
Now put all the given values in this formula, we get the molecular weight of a gas.
Therefore, the molecular weight of a gas is, 128.9 g/mole
<span>This question asksyou to apply Hess's law.
You have to look for how to add up all the reaction so that you get the net equation as the combustion for benzene. The net reaction should look something like C6H6(l)+ O2 (g)-->CO2(g) +H2O(l). So, you need to add up the reaction in a way so that you can cancel H2 and C.
multiply 2 H2(g) + O2 (g) --> 2H2O(l) delta H= -572 kJ by 3
multiply C(s) + O2(g) --> CO2(g) delta H= -394 kJ by 12
multiply 6C(s) + 3 H2(g) --> C6H6(l) delta H= +49 kJ by 2 after reversing the equation.
Then,
6 H2(g) + 3O2 (g) --> 6H2O(l) delta H= -1716 kJ
12C(s) + 12O2(g) --> 12CO2(g) delta H= -4728 kJ
2C6H6(l) --> 12 C(s) + 6 H2(g) delta H= - 98 kJ
______________________________________...
2C6H6(l) + 16O2 (g)-->12CO2(g) + 6H2O(l) delta H= - 6542 kJ
I hope this helps and my answer is right.</span>
