Answer:
The volume in the first tank = 0.32 
The volume in the second tank = 2.066
The final pressure of the mixture = 203.64 K pa
Explanation:
<u>First Tank </u>
Mass = 2 kg
Pressure = 550 k pa
Temperature = 25 °c = 298 K
Gas constant for nitrogen = 0.297 
From the ideal gas equation
P V = m R T
550 × V = 2 × 0.297 × 298
V = 0.32
This is the volume in the first tank.
<u>Second tank</u>
Mass = 4 kg
Pressure = 150 K pa
Temperature = 25 °c = 298 K
Gas constant for oxygen = 0.26
From the ideal gas equation
P V = m R T
150 × V = 4 × 0.26 × 298
V = 2.066
This is the volume in the second tank.
This is the iso thermal mixing. i.e.
----- (1)
Put this value in equation (1)
× 2.386 = 550 × 0.32 + 150 × 2.066
= 203.64 K pa
Therefore the final pressure of the mixture = 203.64 K pa
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
<h3>Further explanation
</h3>
Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:
- 1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.
- 2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product
- 3. Select the coefficient of the substance with the most complex chemical formula equal to 1
For gas combustion reaction which is a reaction of hydrocarbons with oxygen produces CO₂ and H₂O (water vapor). can use steps:
Balancing C atoms, H and the last O atoms
Reaction
Zn + HNO₃⇒ Zn(NO₃)₂ + NO + H₂O
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O
Zn : left = a, right =1 ⇒a=1
H : left = b, right = 2d⇒ b=2d (eq 1)
N : left = b, right = 2+c⇒b=2+c (eq 2)
O : left = 3b, right = 6+c+d ⇒3b=6+c+d(eq 3)
3(2d)=6+c+d
6d=6+c+d
5d=6+c (eq 4)
3(2+c)=6+c+d
6+3c=6+c+d
2c=d (eq 5)
5(2c)=6+c
10c=6+c
9c=6
c = 2/3
d = 2 x 2/3
d = 4/3
b = 2 x 4/3
b = 8/3
The equation
aZn + bHNO₃⇒ Zn(NO₃)₂ + cNO + dH₂O to
Zn + 8/3HNO₃⇒ Zn(NO₃)₂ + 2/3NO + 4/3H₂O x 3
3Zn + 8HNO₃⇒ 3Zn(NO₃)₂ + 2NO + 4H₂O
Answer:
Collisions between gas particles are elastic; there is no net gain or loss of kinetic energy.
Explanation:
When a gas is paced in a container, the molecules of the gas have little or no intermolecular interaction between them. There is a lot of space between the molecules of the gas.
The gas molecules move at very high speed and collide with each other and with the walls of container.
The collision of these particles with each other is perfectly elastic hence the kinetic energy of the colliding gas particles do not change.
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
<h3>What is the boiling-point elevation?</h3>
Boiling-point elevation describes the phenomenon that the boiling point of a liquid will be higher when another compound is added, meaning that a solution has a higher boiling point than a pure solvent.
- Step 1: Calculate the molality of the solution.
We will use the definition of molality.
b = mass solute / molar mass solute × kg solvent
b = 30.0 g / (58.44 g/mol) × 3.75 kg = 0.137 m
- Step 2: Calculate the boiling-point elevation.
We will use the following expression.
ΔT = Kb × m × i
ΔT = 0.512 °C/m × 0.137 m × 2 = 0.140 °C
where
- ΔT is the boiling-point elevation
- Kb is the ebullioscopic constant.
- b is the molality.
- i is the Van't Hoff factor (i = 2 for NaCl).
The normal boiling-point for water is 100 °C. The boiling-point of the solution will be:
100 °C + 0.140 °C = 100.14 °C
Assuming an ebullioscopic constant of 0.512 °C/m for the water, If you add 30.0g of salt to 3.75kg of water, the boiling-point elevation will be 0.140 °C and the boiling-point of the solution will be 100.14 °C.
Learn more about boiling-point elevation here: brainly.com/question/4206205
Yes, they are the same. They are equivalent.
