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gogolik [260]
3 years ago
5

The density of water is about 1.0 g/mL at room temperature. Briefly explain how the density of an aqueous solution at room tempe

rature can be significantly less than 1.0 g/mL. Give an example of such a solution.
Chemistry
1 answer:
maksim [4K]3 years ago
7 0

Answer:

The density of water is about 1.0 g/mL at room temperature.

Briefly explain how the density of an aqueous solution at room temperature can be significantly less than 1.0 g/mL.

Give an example of such a solution.

Explanation:

That means 1.0mL of water weighs ---- 1.0g

If any other aqueous solution which has mass less than 1.0g will have density less than 1.0g.

For example aqueous solution of sulfuric acid has density 0.98g/mL.

That means 1mL of sulfuric acid has mass 0.98g.

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Gloria is making a model of an atom. She uses three different colors to represent the three basic particles that make up the ato
xz_007 [3.2K]

Answer:

protons and neutrons

Explanation:

The nucleus of the atom contains protons and neutrons. The electrons of the atom orbit the nucleus.

8 0
3 years ago
How many grams of potassium (K) contain 5.11 x 10^22 atoms of potassium?
Darina [25.2K]

The atomic mass of K is 39

from Avogadro's law

39g of K contains 6.02x10^23 atoms

therefore if

39=6.02x19^23

X=5.11×10^22

making X the subject of the formula

X= (5.11×10^22×39)÷6.02×10^23

X= 33g

7 0
2 years ago
450g of chromium(iii) sulfate reacts with excess potassium phosphate. How many grams of potassium sulfate will be produced? (ANS
Irina18 [472]

Answer:

600 g K₂SO₄

Explanation:

First write down the complete, balanced chemical equation for such question. In this case:

Cr₂(SO₄)₃ + 2K₃PO₄ →  3K₂SO₄ + 2CrPO₄

Next, calculate molar masses of required compounds mentioned in the question. In this case it is for Cr₂(SO₄)₃ and K₂SO₄.

  • Molar mass(MM) of Cr₂(SO₄)₃:

        = 2*(MM of Cr) + 3*( MM of S) + 3*4*( MM of O)

        = 2*(52) + 3*(32) + 12*(16)

        = 104 + 96 + 192

        = 392 g

  • Molar mass(MM) of K₂SO₄:

        = 2*(MM of K) + 1*( MM of S) + 4*( MM of O)

        = 2*(39) + 32 + 4*(16)

        = 78 + 32 + 64

        = 174 g

Here comes the concept of Limiting reagent:

The limiting reagent in a chemical reaction is the substance that is totally consumed when the chemical reaction is complete. The amount of product formed is limited by this reagent, since the reaction cannot continue without it. The other reactants present with this are usually in excess and called excess reactants. If quantities of both the reactants are given, then one should apply unitary method and find out the limiting reagent out of the two. Then, determine the amount of product formed or percentage yield.

Also, 1 mole( 392 g) of Cr₂(SO₄)₃ gives 3 moles( 174*3 = 522 g) of K₂SO₄.

Using unitary method, if 392g of Cr₂(SO₄)₃ gives 522 g of K₂SO₄ , then 450 g of Cr₂(SO₄)₃ will give how much of K₂SO₄?

Yeild of K₂SO₄ : \frac{522 * 450}{392}

That is 599.3 g.

Since we have not considered molecular masses of individual atoms to 6 decimal places, this number can be approximated to 600g.

Therefore, 600g of K₂SO₄ is produced.

6 0
3 years ago
How many Protons, electrons, and neutrons are in Dysprosium? <br><br> EASY POINTS?
Nadya [2.5K]

Answer:

Number of Protons - 66

Number of Neutrons - 97

Number of Electrons - 66

Explanation:

You're welcome :3

4 0
3 years ago
What is the volume, in cubic meters, of an object that is 0.21 m long, 4.7 m wide, and 5.3 m high?
oee [108]

Answer:

The formula for volume of a rectangle is length multiply by width multiply thus, 0.25 m multiply 6.1 m multiply by 4.9 m = 7.5m^3.

Explanation:

the least number of significant figures is 2 thus the final answer will have the same number of significant figures. 7.5m^3

5 0
3 years ago
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