The volume that will be occupied at 735 torr and 57 c is 23.12 L
<u><em>calculation</em></u>
- <u><em> </em></u> At STP temperature=273 k and pressure=760 torr
- <u><em> </em></u>by use of combined gas formula
that is P1V1/T1= P2V2/T2
where; P1 =760 torr
T1= 273 K
V1= 18.5 L
P2= 735 torr
T2= 57+273= 330 K
V2=?
- by making V2 the formula of subject
V2= T2P1V1/P2T1
V2= [(18.5L x 330 k x 760 torr)/(735 torr x 273 k)]= 23.12 L
Heat transfer is the phenomenon that occurs when the two objects are in the vicinity of each other and by increasing the area of their contact. Thus, option B is correct.
<h3>What is heat transfer?</h3>
Heat transfer is a process that flows the heat from one system to another, and is because of the difference in the temperature of the two objects that are part of the system.
The methods like conduction, convection, and radiation transfer the heat from the surface area to the other object. The heat gets transferred from the area of high to the low temperature.
Therefore, option B. by increasing the surface area the heat transfer increases.
Learn more about heat transfer here:
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Answer:
0.0931 is the ans i think
It is the mass of one mole of the substance.
Answer:

Explanation:
Hello,
In this case, we write the reaction again:

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

Best regards.