The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.
The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.
Thus the water sample of the chemists is both a mixture and pure substance.
Answer:
Empirical formula = K₂SO₄
The name of the compound is Potassium tetraoxosulfate (vi)
Explanation:
Percentage composition of the elements:
Potassium = 44.82%
Sulfur = 18.39%
Oxygen = 36.79%
Mole ratio of the elements:
Potassium = 44.82/39 =1.15
Sulfur = 18.39/32 = 0.57
Oxygen = 36.79/16 = 2.29
Dividing by the smallest ratio in order to obtain the simplest ratio
Potassium = 1.15/0.57 =2
Sulfur = 0.57/0.57 = 1
Oxygen = 2.29/0.57 = 4
Therefore, the mole ratio of Potassium : Sulfur : Oxygen = 2:1:4
Empirical formula = K₂SO₄
The name of the compound is Potassium tetraoxosulfate (vi)
C. Spacecraft are built to be airtight.
Answer:
The fraction of water body necessary to keep the temperature constant is 0,0051.
Explanation:
Heat:
Q= heat (unknown)
m= mass (unknown)
Ce= especific heat (1 cal/g*°C)
ΔT= variation of temperature (2.75 °C)
Latent heat:
ΔE= latent heat
m= mass (unknown)
∝= mass fraction (unknown)
ΔHvap= enthalpy of vaporization (539.4 cal/g)
Since Q and E are equal, we can match both equations:

Mass fraction is:


∝=0,0051