_________ have small sized particles that do not settle out, cannot be filtered out, and do not scatter light.

Chemistry

1 answer:

Naily [24]3 years ago

3 0

Solutions have small sized particles that do not settle out, cannot be filtered out, and do not scatter light.

Colloids have small sized particles that do not settle out and cannot be filtered out, but they scatter light.

You might be interested in

An asteroid hit the Yucatán Peninsula about 65 million years ago,

Morgarella [4.7K]

Answer:

C. It blocked out sunlight

Explanation:

The asteroid impact 65 million years ago was devastating for lot of the animals and plants, resulting in a mass extinction, most noticeably of the dinosaurs. The impact itself killed a lot of flora and fauna, but it was the effect afterward that was crucial. From this impact, there was so much debris propelled into the atmosphere that the sunlight got blocked. Once this happened, most of the plants died out as they were not able to perform photosynthesis. Because the plants were dying out, the herbivores were left without food, so soon they followed, and because the herbivores died out, the predators died out to because they were left without food. The winners turned out to be the small and not specialized plants and animals.

7 0

2 years ago

Explain how water might have two different densities

MrMuchimi

Answer:

I think this is it

Water can have two different densities if it has substances dissolved in it. ... When liquid water freezes it becomes solid water or ice, which is less dense than liquid water. The fact that solid water (ice) is less dense than liquid water is evident in the way ice floats in a glass of water.

Explanation:

6 0

2 years ago

1.) The process for converting ammonia to nitric acid involves the conversion of NH3 to

Firdavs [7]

Answer:

a) 1.39 g ; b) O₂ is limiting reactant, NH₃ is excess reactant; c) 0.7 g

Explanation:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

MM: 17.03 32.00 30.01

4NH₃ + 5O₂ ⟶ 4NO + 6H₂O

Mass/g: 1.5 1.85

2. Calculate the moles of each reactant

$\text{moles of NH}_{3} = \text{1.5 g NH}_{3} \times \dfrac{\text{1 mol NH}_{3}}{\text{17.03 g NH}_{3}} = \text{0.0881 mol NH}_{3}\\\\\text{moles of O}_{2} = \text{1.85 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.057 81 mol O}_{2}$