The molarity of the stock solution is 1.25 M.
<u>Explanation:</u>
We have to find the molarity of the stock solution using the law of volumetric analysis as,
V1M1 = V2M2
V1 = 150 ml
M1 = 0.5 M
V2 = 60 ml
M2 = ?
The above equation can be rearranged to get M2 as,
M2 = 
Plugin the values as,
M2 = 
= 1.25 M
So the molarity of the stock solution is 1.25 M.
Answer:
18.3 kilopascals
Explanation:
We are given that the volume of this container is 0.0372 meters^3, that the mass of water is 4.65 grams, and that the temperature of this water vapor ( over time ) is 368 degrees Kelvins. This is a problem where the ideal gas law is an " ideal " application.
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First calculate the number of moles present in the water ( H2O ). Water has a mass of 18, so it should be that n, in the ideal gas law - PV = nRT, is equal to 4 / 18. It is the amount of the substance.
We now have enough information to solve for P in PV = nRT,
P( 0.0372 ) = 4 / 18( 8.314 )( 368 ),
P ≈ 18,276.9
Pressure ≈ 18.3 kilopascals
<u><em>Hope that helps!</em></u>
When the pressure is increased, the equilibrium will shift to the left to offset the pressure increase. Equilibrium shifting to the left side is favored because the left side has fewer moles of gas than the number of moles of gas on the right-hand side and because it exerts less pressure. Therefore, the answer is there will be a shift toward the reactants.
Answer:
The soil has a percentage of water by mass of 27.8 %.
Explanation:
Keeping in mind that
- Mass of Clay = Mass of water + Mass of Dry Soil
we can <u>calculate the mass of water</u>:
- 1350 g = Mass of Water + 975 g
We can then <u>calculate the mass percentage of water in the soil</u>:
- 375 / 1350 * 100% = 27.8 %
C - bc as the water boils the cold water particles melt and become the hot water particles!
