An increase in temperature will increase the average kinetic energy of the molecules. As the particles move faster, they will likely hit the edge of the container more often.
Q=m°C<span>ΔT
=(500g) x (1 cal/g.</span>°C) x (48°C-21°C) = 13500 cal
13500 cal / 1000 = 13.5 kcal
<span>"What is the caloric value (kcal/g) of the french fries?"
13.5 kcal/ 2.5 g = 5.4 kcal/g</span>
Answer: A. Sun, Moon, Earth
Answer: 2.71 moles of solute for every 1 kg of solvent.
Explanation: As you know, the molality of a solution tells you the number of moles of solute present for every 1 kg of the solvent.This means that the first thing that you need to do here is to figure out how many grams of water are present in your sample. To do that, use the density of water.500.mL⋅1.00 g1mL=500. g Next, use the molar mass of the solute to determine how many moles are present in the sample.115g⋅1 mole NanO385.0g=1.353 moles NaNO3So, you know that this solution will contain 1.353moles of sodium nitrate, the solute, for 500. g of water, the solvent.In order to find the molality of the solution, you must figure out how many moles of solute would be present for 1 kg=103g of water.103g water⋅1.353 moles NaNO3500.g water=2.706 moles NaNO3You can thus say that the molality of the solution is equal to molality=2.706 mol kg−1≈2.71 mol kg−1 The answer is rounded to three sig figs.
Answer:
0.88 g
Explanation:
Using ideal gas equation to calculate the moles of chlorine gas produced as:-
where,
P = pressure of the gas = 805 Torr
V = Volume of the gas = 235 mL = 0.235 L
T = Temperature of the gas = ![25^oC=[25+273]K=298K](https://tex.z-dn.net/?f=25%5EoC%3D%5B25%2B273%5DK%3D298K)
R = Gas constant = 
n = number of moles of chlorine gas = ?
Putting values in above equation, we get:
According to the reaction:-
1 mole of chlorine gas is produced when 1 mole of manganese dioxide undergoes reaction.
So,
0.01017 mole of chlorine gas is produced when 0.01017 mole of manganese dioxide undergoes reaction.
Moles of 
= 0.01017 moles
Molar mass of = 86.93685 g/mol
So,
Applying values, we get that:-
<u>0.88 g of 
should be added to excess HCl (aq) to obtain 235 mL of 
at 25 degrees C and 805 Torr.</u>
