Whose measurements are more precise?20.3 cm and 21.0cm


Answer 1


will i think the 21.0cm

Answer 2
Answer: 20.3 because it isn’t rounded

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Practice the Skill 21.15b When the following ketone is treated with aqueous sodium hydroxide, the aldol product is obtained in poor yields. In these cases, special distillation techniques are used to increase the yield of aldol product. Predict the aldol addition product that is obtained, and propose a mechanism for its formation. For the mechanism, draw the curved arrows as needed. Include lone pairs and charges in your answer. Do not draw out any hydrogen explicitly in your products. Do not use abbreviations such as Me or Ph.



Check the explanation


Kindly check the attached image below for the step by step explanation to the question above.

In an attempt to prepare n−propylbenzene, a chemist alkylated benzene with 1−chloropropane and aluminum chloride. However, two isomeric hydrocarbons were obtained in a ratio of 2:1, the desired n−propylbenzene being the minor component. What do you think was the major product? How did it arise?



2-Phenylpropane (Cumene)


Famous Friedel Craft Alkylation.

Aluminum Chloride grasped the 1-chloropropane forming an intermediate product composed of Aluminum tetrachloride and n-propylcation. It is well understood that primary carbocations are unstable and therefore undergo hydrogen shifting to attain stability. n-Propylcarbocation undergone hydrogen shifting, forming isopropylcarbocation which reacted with benzene forming 2-Phenylpropane as the major product and HCl as a byproduct.

AlCl3 + CH3CH2CH2Cl --> AlCl4-   +   CH3CH2CH2+

CH3CH2CH2+  ---> CH3CH(+)CH3

C6H6 + CH3CH(+)CH3 ---> C6H5CH(CH3)2   + H+

AlCl4- + H+ ---> HCl + AlCl3


From the given problem statement,he was attempting to prepare n−propylbenzene by alkylation benzene with 1−chloropropane and aluminum chloride,but 1-propyle benze was a major product in result.

As the speed of the particles decreases, -a. Intermolecular forces become stronger

b. Intermolecular forces become weaker

c. Intermolecular forces do not change

d. Energy increases


As the speed of the particles decreases, intermolecular forces become stronger. Thus, the correct option for this question is A.

What are Intermolecular forces?

The intermolecular forces may be defined as the forces of attraction. that is present between atoms, molecules, and ions when they are placed close to each other in order to form a compound or element. This force is continuously acting on the neighboring particles of different molecules.

It is found that at low temperature, when the speed of molecules/particles decrease, they migrate closer to one another that results in their intermolecular forces that become stronger as compared to the initial one.

As the attraction between molecules gradually increases, their movement decreases, and undergo fewer collisions between them.

Therefore, as the speed of the particles decreases, intermolecular forces become stronger. Thus, the correct option for this question is A.

To learn more about Intermolecular forces, refer to the link:






How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?


0.085 moles of Al are required to form 23.6 g of AlBr₃.

Let's consider the following balanced equation for the synthesis reaction of AlBr₃.

2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s)

First, we will convert 23.6 g to moles using the molar mass of AlBr₃ (266.69 g/mol).

23.6 g * (1mol)/(266.69g) = 0.0885 mol

The molar ratio of Al to AlBr₃ is 2:2. The moles of Al required to form 0.0885  moles of AlBr₃ are:

0.0885molAlBr_3 * (2molAl)/(2molAlBr_3) = 0.0885molAl

0.085 moles of Al are required to form 23.6 g of AlBr₃.

You can learn more about stoichiometry here: brainly.com/question/22288091


0.088 mole of Al.


First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

Oxygen is a __________ and nitrogen is a __________. metalloid, metalloid nonmetal, metal nonmetal, nonmetal nonmetal, metalloid metal, metalloid



"nonmetal, nonmetal"


Oxygen is a non metal and Nitrogen is a non metal. It is 8th element of the periodic table. It is located in period 2 and group 16.

Nitrogen lies at the group 15 of the periodic table. Its atomic no is 7. Its valency is 2.

Hence, the correct option is (c) "nonmetal, nonmetal".

Final answer:

Oxygen and nitrogen are both nonmetals. They are unable to conduct heat or electricity effectively and are typically found on the right side of the periodic table.


In the periodic table of elements, oxygen and nitrogen are both classified as nonmetals. Nonmetals are elements that are not able to conduct electricity or heat very well. As opposed to metals, nonmetals are brittle and do not have the ability to be shaped into thin sheets or wires. They are typically found on the right side of the periodic table and are represented by groups 14-17. So, to answer the student's question, oxygen is a nonmetal and nitrogen is a nonmetal.

Learn more about Nonmetals here:



The specific heat capacity of liquid water is 4.18 J/g-K. How many joules of heat are needed to raise the temperature of 5.00 g of water from 25.1°C to 65.3°C?





Use the equation: Q = mcΔT

m = mass (5 g)

c = specific heat (4.18)

ΔT = change in temperature (65.3-25.1 = 40.2)

5*4.18*40.2 = 840.12