The initial temperature is 137.34 °C.
<u>Explanation:</u>
As the specific heat formula says that the heat energy required is directly proportional to the mass and change in temperature of the system.
Q = mcΔT
So, here the mass m is given as 23 kg, the specific heat of steel is given as c = 490 J/kg°C and the initial temperature is required to find with the final temperature being 140 °C. Also the heat energy required is 30,000 J.
ΔT =
ΔT =
Since the difference in temperature is 2.66, then the initial temperature will be
Final temperature - Initial temperature = Change in temperature
140-Initial temperature = 2.66
Initial temperature = 140-2.66 = 137.34 °C
Thus, the initial temperature is 137.34 °C.
Molar mass is the ratio of the mass to that amount of the substance. The mass of the barium nitrate in the formula unit is 23.0 grams.
<h3>What is mass?</h3>
The mass of a substance is the product of the molar mass of the compound and the number of moles of the compound.
Given,
Molar mass of barium nitrate = 261.35 g/mol
If,
have a mass of 261.35 g/mol then,
formula units will have a mass of,

Therefore, option C. 23.0 gm is the mass of barium nitrate.
Learn more about mass here:
brainly.com/question/24958554
Answer:
C.)house temperature increases cause the bimetallic coil to expand
Explanation:
A bimetallic strip is used in an air conditioner as a regulator. It is made up of two metals that behaves differently to a change in temperature.
To turn on the air conditoner, the circuit must be complete. When the temperature is high in a room, bimetallic strip responds by expanding and bending to complete the circuit. This puts on the air conditioner and the house begins to cool through exchange of warm and cold air.
When the temperature is too cold, the strip contracts and straightens. This process disconnects the circuit and the air conditioning system ceases to work.
22.4L
of any gas contains 1 mol of that gas.
50.75g/10L*22.4L/1 mol= 113.68g/mol- this is the mole weight of your gas
1 mol/113.68g*129.3g=1.137403 mol
Set up a ratio
1.137403mol/x L=1 mol/22.4 L
X=25.477827L, or with sig figs, x=25.5L