Answer:
THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.
AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.
Explanation:
The question above follows Boyle's law of the gas law as the temperature is kept constant.
Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.
Mathematically, P1 V1 = P2 V2
P1 = 150 kPa = 150 *10^3 Pa
V1 = 2.48 L
V2 = 2.98 L
P2 = ?
Rearranging the equation, we obtain;
P2 = P1 V1 / V2
P2 = 150 kPa * 2.48 / 2.98
P2 = 372 *10 ^3 / 2.98
P2 = 124.8 kPa.
The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.
Answer:
(See explanation for further details)
Explanation:
1) The quantity of moles of sulfur is:


2) The number of atoms of helium is:


3) The quantity of moles of carbon monoxide is:


4) The number of molecules of sulfur dioxide is:


5) The quantity of moles of sodium chloride is:


6) The number of formula units of magnesium iodide is:


7) The quantity of moles of potassium permanganate is:


8) The number of molecules of carbon tetrachloride is:


9) The quantity of moles of aluminium is:


10) The number of molecules of oxygen difluoride is:


Answer:
Acid solution
Explanation:
In acid-base titrations carried out in school and college labs, many of the older generation, well-informed teachers told their students that the acid solution should be taken in the burette.
When you use exponents, think of it like this. 1 squared is 1 x 1. 1 cubed is 1 x 1 x 1. And 1 to the power of 4 is 1 x 1 x 1 x 1. And so on. You basically just multiply them by themselves. 0 to the power of any exponent is 1, well that's just a rule. It doesn't make much sense but it's easy to remember and I wouldn't worry about it.
Answer:
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
Explanation:
One colligative property is the freezing point depression due the addition of a solute. The equation is:
ΔT=Kf*m*i
<em>Where ΔT is change in temperature = 0.400°C</em>
<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>
<em>m is molality of the solution (Moles of solute / kg of solvent)</em>
<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>
Replacing:
0.400°C =1.86°C/m*m*1
0.400°C / 1.86°C/m*1 = 0.215m
As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:
0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.
The mass of ethylene glycol must be added is:
0.0602 moles * (62.10g / mol) =
3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C
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