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Katen [24]
3 years ago
12

How many divisions are used to classify living organisms? a) 5 b) 6 c) 8 d) 7

Chemistry
1 answer:
Nataly_w [17]3 years ago
4 0
Answer:a

Living things are divided into five kingdoms: animal, plant, fungi, protist and monera
You might be interested in
I need an example of a chemical change involving water.Pleaseeee.
slava [35]
Chemical change occur when two substances are combined and produces a new substance or decomposes into two or more substances which are entirely different from the original two substances.

There are three types of chemical changes. These are 1) Inorganic Changes, 2) Organic Changes, and 3) Biochemical Changes


Here are some examples of chemicsal changes.

If you combine Sodium and Water, chemical changes causes decomposition into Sodium Hydroxide and Hydrogen.

Sodium + Water  ==>  Sodium Hydroxide and Hydrogen

Na + H2O ====>  NaOH and H

Another example of chemical change is:

Carbon Dioxide and Water will decompose into Sugar and Oxygen

Carbon Dioxide + Water ==> Sugar and Oxygen

CO2 + H2O ==> CnH2nOn (where n is between 3 and 7) and O

 
7 0
3 years ago
Briefly describe the intermolecular forces present in your unknown
Korvikt [17]

An inter-molecular power is basically an alluring power between neighboring particles. There are three regular sorts of inter-molecular power: lasting dipole-dipole powers, hydrogen bonds and van der Waals' powers.

7 0
3 years ago
Which element is found in period 6 group 14<br>​
kicyunya [14]

Answer:

Lead

that can be found in the periodic table of elements

Explanation:

7 0
3 years ago
The ksp of yttrium fluoride, yf3, is 8.62 × 10-21. calculate the molar solubility of this compound.
motikmotik

Answer:

The molar solubility of YF₃ is 4.23 × 10⁻⁶ M.

Explanation:

In order to calculate the molar solubility of YF₃ we will use an ICE chart. We identify 3 stages: Initial, Change and Equilibrium and we complete each row with the concentration of change of concentration. Let's consider the solubilization of YF₃.

       YF₃(s) ⇄ Y³⁺(aq) + 3 F⁻(aq)

I                       0               0

C                     +S            +3S

E                       S              3S

The solubility product (Ksp) is:

Ksp = [Y³⁺].[F⁻]³= S . (3S)³ = 27 S⁴

S=\sqrt[4]{Ksp/27} =\sqrt[4]{8.62 \times 10^{-21}  /27}=4.23 \times 10^{-6}M

6 0
3 years ago
A chemistry student needs 60.0 g of 2-ethyltoluene for an experiment. He has available 1.5 kg of a 15.3% w/w solution of 2 ethyl
Mila [183]

Answer:

392 g

Explanation:

The given concentration tells us that<em> in 100 g of solution, there would be 15.3 g of 2-ethyltoluene</em>.

With that in mind we can<u> calculate how many grams of solution would contain 60.0 g of 2-ethyltoluene</u>:

  • Mass of solution * 15.3 / 100 = 60.0 g 2-ethyltoluene
  • Mass of solution = 392 g
8 0
3 years ago
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