CHEMISTRY HIGH SCHOOL

Which of the following is a molecular compound?F) NO2
G) Ca(NO3)2
H) H2SO4
J) NaOH

Answers

Answer 1
Answer:

Answer: H

Explanation:
Answer 2
Answer: The answer is G my guy

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Which pair of elements is MOST likely to chemically combine and form ionic bonds?. A). chlorine and chlorine. B). lithium and chlorine. C). carbon and hydrogen. D). sulfur and oxygen

Answers

Answer: B). lithium and chlorine

Explanation: An ionic bond is formed between a metal and a non-metal by transfer of electrons from metal to non metal.

Electronic configuration of lithium:

[Li]:31s^22s^1

Lithium atom will loose one electron to gain noble gas configuration and form lithium cation with +1 charge.

[Li^+]=1s^22s^2

Chlorine atom has oxidation number of -1.

Electronic configuration of chlorine;

[Cl]=1s^22s^22p^63s^23p^5

Chlorine atom will gain one electron to gain noble gas configuration and form chloride ion with -1 charge.

[Cl^-]=1s^22s^22p^63s^23p^6

Li^+ combine with Cl^- to form lithium chloride (LiCl)
The pair of elements that will most likely combine chemically to form ionic bonds are B. Lithium and Chlorine.

Using the periodic table, determine which material is most likely to be a good insulator. magnesium silver aluminum sulfur

Answers

The correct answer is this one: "Magnesium." Using the periodic table, the material that is most likely to be a good insulator is the magnesium. Magnesium is a chemical element with symbol Mg and atomic number 12. It is a shiny gray solid which bears a close physical resemblance to the other five elements in the second column of the periodic.

Using the periodic table, determine which material is most likely to be a good insulator.

  1. magnesium
  2. silver
  3. aluminum
  4. sulfur

the correct answer is 4. sulfur

Write the name of each of the following Ionic Compounds: Ba3(PO4)2 , MgSO4 , PbO2

Answers

1. Barium phosphate
2. Magnesium sulphate
3. Lead(IV) oxide

Final answer:

Ba3(PO4)2 is Barium Phosphate, MgSO4 is Magnesium Sulfate, and PbO2 is Lead (IV) Oxide or Plumbic Oxide.

Explanation:

The Ionic Compounds you have mentioned are named as follows:

  • Ba3(PO4)2 is named Barium Phosphate.
  • MgSO4 is named Magnesium Sulfate.
  • PbO2 is named Lead (IV) Oxide or Plumbic Oxide. Always remember for ionic compounds containing transition metals (like Pb), we have to indicate the charge of the cation.

Learn more about Ionic Compounds here:

brainly.com/question/33500527

The ancient philosophers discussed the world around thembut did not do any experiments or detailed observations.
TRUE
FALSE

Answers

Answer:

FALSE

Explanation:

Final answer:

The claim that ancient philosophers did not conduct experiments or detailed observations is false. Philosophers like Aristotle made extensive contributions to biology through observation, and Hellenistic scholars conducted early forms of experimental research.

Explanation:

The statement that ancient philosophers discussed the world around them but did not do any experiments or detailed observations is FALSE. While it is true that some ancient philosophers, particularly in the early stages of Greek philosophy, were more speculative in nature, others, such as Aristotle, engaged in detailed observations and can be considered early scientists. For example, Aristotle conducted extensive studies on biology, classifying a vast number of species and making observations on their life cycles, habits, and anatomy.

Throughout history, methodologies have varied, and while the ancient philosophers didn't conduct experiments in the modern sense, they did employ various forms of inquiry to understand the world. Furthermore, in the Hellenistic period, scholars at institutions like the Library of Alexandria did indeed conduct forms of research that could be seen as experimental, such as Eratosthenes estimating the Earth's circumference.

Learn more about Ancient Philosophy and Science here:

brainly.com/question/12278195

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What would happen if a decomposers were absent from the forest ecosystem?

Answers

there would be no way that the nutrients could go back into the ecosystem and there would be tons of dead animals lying around.

An automobile antifreeze mixture is made by mixing equal volumes of ethylene glycol (d = 1.114 g/mL; M = 62.07 g/mol) and water (d = 1.00 g/mL) at 20°C. The density of the mixture is 1.070 g/mL. Express the concentration of ethylene glycol as (a) volume percent 50 % v/v (b) mass percent 52.7 % w/w (c) molarity M (d) molality m (e) mole fraction

Answers

Answer :

(a) The volume percent is, 50.63 %

(b) The mass percent is, 52.69 %

(c) Molarity is, 9.087 mole/L

(d) Molality is, 17.947 mole/L

(e) Moles fraction of ethylene glycol is, 0.244

Explanation : Given,

Density of ethylene glycol = 1.114 g/mL

Molar mass of ethylene glycol = 62.07 g/mole

Density of water = 1.00 g/mL

Density of solution or mixture = 1.070 g/mL

According to the question, the mixture is made by mixing equal volumes of ethylene glycol and water.

Suppose the volume of each component in the mixture is, 1 mL

First we have to calculate the mass of ethylene glycol.

\text{Mass of ethylene glycol}=\text{Density of ethylene glycol}* \text{Volume of ethylene glycol}=1.114g/mL* 1mL=1.114g

Now we have to calculate the mass of water.

\text{Mass of water}=\text{Density of water}* \text{Volume of water}=1.00g/mL* 1mL=1.00g

Now we have to calculate the mass of solution.

Mass of solution = Mass of ethylene glycol + Mass of water

Mass of solution = 1.114 + 1.00 = 2.114 g

Now we have to calculate the volume of solution.

\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=(2.114g)/(1.070g/mL)=1.975mL

(a) Now we have to calculate the volume percent.

\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}* 100=(1mL)/(1.975mL)* 100=50.63\%

(b) Now we have to calculate the mass percent.

\text{Mass percent}=\frac{\text{Mass of ethylene glycol}}{\text{Mass of solution}}* 100=(1.114g)/(2.114g)* 100=52.69\%

(c) Now we have to calculate the molarity.

\text{Molarity}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Volume of solution (in mL)}}

\text{Molarity}=(1.114g* 1000)/(62.07g/mole* 1.975L)=9.087mole/L

(d) Now we have to calculate the molality.

\text{Molality}=\frac{\text{Mass of ethylene glycol}* 1000}{\text{Molar mass of ethylene glycol}* \text{Mass of water (in g)}}

\text{Molality}=(1.114g* 1000)/(62.07g/mole* 1kg)=17.947mole/kg

(e) Now we have to calculate the mole fraction of ethylene glycol.

\text{Mole fraction of ethylene glycol}=\frac{\text{Moles of ethylene glycol}}{\text{Moles of ethylene glycol}+\text{Moles of water}}

\text{Moles of ethylene glycol}=\frac{\text{Mass of ethylene glycol}}{\text{Molar of ethylene glycol}}=(1.114g)/(62.07g/mole)=0.01795mole

\text{Moles of water}=\frac{\text{Mass of water}}{\text{Molar of water}}=(1g)/(18g/mole)=0.0555mole

\text{Mole fraction of ethylene glycol}=(0.01795mole)/(0.01795mole+0.0555mole)=0.244
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