It is a easy one. Keep working and don't stress about it, you'll see, doing this problem alone will securise you
Answer:
The moles of KClO3 = 0.052 moles
Explanation:
Step 1: Calculate the pressure of oxygen gas
The oxygen has a total pressure (including water vapour) of 760 mmHg
The pressure of Oxygen = (760 - 26) mmHg
= 734 mmHg of water vapor
Step 2: Calculate the no of moles of oxygen
Using Ideal gas equation
P V = n R T
P = pressure of oxygen in N/m2 ( you should convert 734 mmHg to pascal or N/m2) = 97,858.6 N/m2 or pas
V = 2 litres = 0.002 m3
R = gas constant = 8.31
T= 27oC = 300 K
Applying this equation P V = n R T
97,858.6 x 0.002 = n x 8.31 x 300
n = 0.0785 mol of Oxygen
From the balanced equation
2 KClO 3 ---- 2 KCl + 3 O 2
3 moles of oxygen is produced from 2 moles KClO3
so 0.0785 mole of oxygen will be produced from x
x = (0.0785 x 2 ) / 3
x = 0.052 moles of KClO3
Answer:
A. Substance E
A. Substance C
A. Substance A
Explanation:
Given that:
At 4 °C, Substance E has a vapor pressure of 86. torr and Substance F has a vapor pressure of 136. torr
Which has a higher boiling point?
A. Substance E
B. Substance F
C. Neither,EandF have the same boiling point
The vapor pressure varies inversely proportional to the boiling point.
Therefore, the lower the vapor pressure, the higher the boiling point.
At 4°C, Substance E with a lower vapor pressure of 86. torr will have a higher boiling point from the given information.
2.
Recall that :
therefore, the lower the enthalpy of vaporization, the higher the vapor pressure at any given temperature.
Given that:
Substance C has an enthalpy of vaporization smaller than that of substance D. Then, substance C has a higher vapor pressure.
3.
We've earlier said that:
The vapor pressure varies inversely proportional to the boiling point.
Therefore, the lower the vapor pressure, the higher the boiling point.
As such, Substance A will have a higher boiling point.
That’s easy it would be B.
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.
