Answer:
NaNO3 (solubility = 89.0 g/100 g H2O)
Explanation:
The solubility of a specie is the amount of solute that will dissolve in one litre of the solvent. Solubility is usually expressed in units of molarity.
Now let us calculate the molarity of the NaNO3 (solubility = 89.0 g/100 g H2O)
Molar mass of NaNO3= 23+14+3(16)= 85gmol-1
Mass of solute=89.0g
Amount of solute= mass of NaNO3/molar mass of NaNO3
Amount of solute= 89.0g/85.0 gmol-1
= 1.0moles of NaNO3
Note that 100g of water=100cm^3 of water.
If 1.0 moles of NaNO3 dissolve in 100cm^3 or water therefore,
x moles of NaNO3 will dissolve in 1000cm^3 of water
x= 1.0 × 1000/ 100
x= 10.0 moles of NaNO3
Which of the following is not an example of a mass movement?
to. a. Ground slides b) acid rain c. mud flow d. drop
Answer:
b) acid rain
Explanation:
Acid rain is not an example or a type of mass movement.
Mass movement or mass wasting is the movement of rocks, soils and other debris downslope under the influence of gravity.
Water, slope and gravity are triggers of mass movement.
- Acid rain is a not a form of mass movement
- It occurs with some certain gases mixes rain water to produce an acidic precipitation.
- When the water falls to the each surface, it causes problems to the ecosystem and can also corrode building materials.
Answer:
An intensive property is a property of matter that does not change as the amount of matter changes. It is a bulk property, which means it is a physical property that is not dependent on the size or mass of a sample.
In contrast, an extensive property is one that does depend on sample size. Examples of extensive properties include mass and volume.
Answer:
The dilution factor of protein in tube # 4 is 125. Molar concentration is 0.0088 M protein
Explanation:
The dilution factor indicates how many times is more concentrated a main solution in relationship with a diluted solution. In this case, the main solution is in tube #1. For calculating the dilution factor and molar concentration in tube #4 we need the main solution concentration which comes from next equation:
Initial volume * initial concentration = final volume * final concentration
0.5 mL * 10M = 5mL * final concentration
1.1 M = final concentration = main solution concentration
Applying the same equation for remain tubes we have 0.22 M for tube #2, 0.044 M for tube # 4 and 0.0088 for tube # 4.
Dilution factor = Main solution concentration/tube 4 concentration
Dilution factor = 1.1/0.0088 = 125
I hope my answer helps you
