When cooled by liquid nitrogen, the balloon shrinks (not as much as the air-filled balloon) and it sinks down on the table. When heating up, the balloon slowly rises and flies up in the air again. Explanation 1: The volume of the balloon decreases by the low temperature, because the gas inside is cooled down.
Answer:
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Explanation:
<u>Step 1:</u> Data given
ΔH∘=20.1 kJ/mol
ΔS is 45.9 J/K
<u>Step 2:</u> When is the reaction spontaneous
Consider temperature and pressure = constant.
The conditions for spontaneous reactions are:
ΔH <0
ΔS > 0
ΔG <0 The reaction is spontaneous at all temperatures
ΔH <0
ΔS <0
ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)
ΔH >0
ΔS >0
ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)
<u>Step 3:</u> Calculate the temperature
ΔG <0 = ΔH - T*ΔS
T*ΔS > ΔH
T > ΔH/ΔS
In this situation:
T > (20100 J)/(45.9 J/K)
T > 437.9 K
T > 164.75 °C
The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous
Answer:
0.9612 g
Explanation:
First we <u>calculate how many moles are there in 3.00 g of CCl₃F</u>, using its <em>molar mass</em>:
- 3.00 g CCl₃F ÷ 137.37 g/mol = 0.0218 mol CCl₃F
Now, we need to calculate how many grams of N₂O would have that same number of molecules, or in other words, <em>the same amount of moles</em>.
Thus we <u>calculate how many grams would 0.0218 moles of N₂O weigh</u>, using the <em>molar mass of N₂O</em> :
- 0.0218 mol N₂O * 44.013 g/mol = 0.9612 g N₂O
Answer:
1.84 L
Explanation:
Using the equation for reversible work:
Where:
W is the work done (J) = -287 J.
Since the gas did work, therefore W is negative.
P is the pressure in atm = 1.90 atm.
However, work done is in joules and pressure is in atm. We can use the values of universal gas constant as a convenient conversion unit. R = 8.314 J/(mol*K); R = 0.0821 (L*atm)/(mol*K)
Therefore, the conversion unit is 0.0821/8.314 = 0.00987 (L*atm)/J
is the initial volume = 0.350 L
is the final volume = ?
Thus:
(-287 J)*0.00987 (L*atm)/J = -1.9 atm*( - 0.350) L
= [(287*0.00987)+(1.9*0.350)]/1.9 = (2.833+0.665)/1.9 =1.84 L
Answer:
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
Explanation:
When we balance a chemical equation, what we are trying to do is to achieve the same number of atoms for each element on both sides of the arrow. On the right of the arrow is where we can find the products, while the reactants are found on the left of the arrow.
We usually balance O and H atoms last.
AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 1
Cl --- 3
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
2 AsCl₃ + H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 2
S --- 1
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of As atoms is now balanced.
2 AsCl₃ + 3 H₂S → As₂S₃ +HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 1
H --- 1
S --- 3
The number of S atoms is now equal on both sides.
2 AsCl₃ + 3 H₂S → As₂S₃ + 6 HCl
<u>reactants</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
<u>products</u>
As --- 2
Cl --- 6
H --- 6
S --- 3
The equation is now balanced.
