Answer:
THE PRESSURE OF THE TIRE ON THE TRIP HOME AT THE ROAD SURFACE TEMPERATURE OF 32°C IS 160 kPa.
Explanation:
Initial Pressure = 75 kPa
Initial temperature = 15 °C
Final temperature = 32 °C
Final pressure = unknown
Using the combined equation of gases;
P1V1/T1 = P2V2/ T2
Since the tire will have the same volume of air in it showing that volume of constant both at the repair shop and on the road surface.
The relationship between pressure and temperature is used with constant volume.
P1/T1 = P2/ T2
75 kPa / 15 °C = P2 / 32 °C
P2 = 75 kPa * 32 °C / 15 °C
P2 = 2400 kPa °C / 15 °C
P2 = 160 kPa.
So therefore, the pressure of the tire on the trip home when the temperature of the road surface is 32°C is 160 kPa.
Explanation:
Equation of reaction:
PCl₅ ⇆ PCl₃ + Cl₂
Problem: what direction will the reaction shift towards if PCl₃ is added;
Solution:
According to Le Chatelier's principle "if any conditions of a system in equilibrium is changed, the system will adjust itself in order to annul the effect of the change".
If the concentration of PCl₃ is increase, the equilibrium will shift towards the left because it is used up in that direction. More of the PCl₅ will be produced in order for the system to adjust back to equilibrium.
learn more:
Equilibrium brainly.com/question/11080417
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Explanation:
Carbon dioxide dissolves in water and slowly reacts with water to produce carbonic acid. The cloudy white solution observed when CO2 is bubbled into limewater results from a reaction between Ca(OH)2 and either CO2 or H2CO3 to form an insoluble calcium carbonate precipitate.
Answer:
1279 °C
Explanation:
Data Given:
Amount of Heat absorb = 5.82 x 10³ KJ
Convert KJ to J
1 KJ = 1000 J
5.82 x 10³ KJ = 5.82 x 10³ x 1000 = 5.82 x10⁶ J
mass of sample = 8.92 Kg
Convert Kg to g
1 kg = 1000 g
8.92 Kg = 8.92 x 1000 = 8920 g
Cs of steel = 0.51 J/g °C
change in temperature = ?
Solution:
Formula used
Q = Cs.m.ΔT
rearrange the above equation to calculate the mass of steel sample
ΔT = Q / Cs.m .... . . . . . (1)
Where:
Q = amount of heat
Cs = specific heat of steel = 0.51 J/g °C
m = mass
ΔT = Change in temperature
Put values in above equation 1
ΔT = 5.82 x10⁶ J / 0.51 (J/g °C) x 8920 g
ΔT = 5.82 x10⁶ J /4549.2 (J/°C)
ΔT = 1279 °C
So,
change in temperature = 1279 °C
Answer:
1.13 × 10⁶ g
Explanation:
Let's consider the reduction of aluminum (III) from Al₂O₃ to pure aluminum.
Al³⁺ + 3 e⁻ → Al
We can establish the following relations:
- 1 Ampere = 1 Coulomb / second
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 mole of Al is produced when 3 moles of electrons circulate
- The molar mass of Al is 26.98 g/mol.
The mass of aluminum produced under these conditions is:
