I think it’s Object 3
Hopefully this helps
Answer : The temperature of the air in the tire is, 341 K
Explanation :
Gay-Lussac's Law : It is defined as the pressure of the gas is directly proportional to the temperature of the gas at constant volume and number of moles.

or,

where,
= initial pressure = 198 kPa
= final pressure = 225 kPa
= initial temperature = 
= final temperature = ?
Now put all the given values in the above equation, we get:


Therefore, the temperature of the air in the tire is, 341 K
Answer: There are now 2.07 moles of gas in the flask.
Explanation:
P= Pressure of the gas = 697 mmHg = 0.92 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = ?
n = number of moles = 1.9
T = Temperature of the gas = 21°C=(21+273)K= 294 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
When more gas is added to the flask. The new pressure is 775 mm Hg and the temperature is now 26 °C, but the volume remains same.Thus again using ideal gas equation to find number of moles.
P= Pressure of the gas = 775 mmHg = 1.02 atm (760 mmHg= 1 atm)
V= Volume of gas = volume of container = 49.8 L
n = number of moles = ?
T = Temperature of the gas = 26°C=(26+273)K= 299 K (0°C = 273 K)
R= Value of gas constant = 0.0821 Latm\K mol
Thus the now the container contains 2.07 moles.
Answer:
I dont know the answer for that question it's hard question isn't it
Answer:
The granite block transferred <u>4080 joules</u> of energy, and the mass of the water is <u>35.84 grams</u>.
Explanation:
The equation needed to answer both parts of the question is:
Q = mcΔT
In this equation,
-----> Q = energy/heat (J)
-----> m = mass (g)
-----> c = specific heat (J/g°C)
-----> ΔT = change in temperature (°C)
<u>Part #1:</u>
First, you need to find the energy transferred from granite block using the previous equation. You have been given the mass, specific heat, and change in temperature.
Q = ? J c = 0.795 J/g°C
m = 126.1 g ΔT = 92.6 °C - 51.9 °C = 40.7 °C
Q = mcΔT
Q = (126.1 g)(0.795 J/g°C)(40.7 )
Q = 4080
<u>Part #2:</u>
Secondly, using the energy calculated in Part #1, you need to calculate the mass of the water. You have calculated the energy transferred, and have been given the specific heat and change in temperature.
Q = 4080 J c = 4.186 J/g°C
m = ? g ΔT = 51.9 °C - 24.7 °C = 27.2 °C
Q = mcΔT
4080 J = m(4.186 J/g°C)(27.2 °C)
4080 J = m(113.8592)
35.84 = m