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lora16 [44]
3 years ago
8

Explain the difference between elements and compounds, a teacher filled

Chemistry
1 answer:
stiv31 [10]3 years ago
6 0

Answer:

Kindly check explanation.

Explanation:

So, without mincing words the solution to this particular Question is given below:

First, an ELEMENT is a that entity that has what is known as the atoms. Although, there might be more elements but the recognized elements are group into the periodic table/chart. Examples of elements are Hydrogen, Helium, Lithium, Beryllium, Boron, carbon, Nitrogen, oxygen, Fluorine, neon, sodium, magnesium, Aluminium and so on.

A COMPOUND consist of different elements that are chemically combined. Examples of compounds are; H2O, CaCO3, H2SO4, HNO3, MgCl2, NaCl, HCl and many more.

Also, elements has the same atoms while componds have different number of atoms since it is a combination of elements.

Finally, ELEMENTS can not be broken down through the use of chemical process while COMPONDS can be broken down through chemical processes.

To the second part of the question. The box that will represent elements will contain balls of the same colors and the balls are not joined together.

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Weathering breaks down rock and other material

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Calculate the number of moles of sodium carbonate originally present, if 6.0 ml of 0.100 M HCl were used to titrate the mixture
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Answer:

0.0006 mole

Explanation:

Considering:

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Or,

Moles =Molarity \times {Volume\ of\ the\ solution}

Given :

For HCl :

Molarity = 0.100 M

Volume = 6.0 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 6.0×10⁻³ L

Thus, moles of potassium iodide :

Moles=0.100 \times {6.0\times 10^{-3}}\ moles

Moles of HCl = 0.0006 moles

From the reaction shown below:-

HCl+Na_2CO_3\rightarrow NaHCO_3+NaCl

1 mole of HCl reacts with 1 mole of sodium carbonate.

So,

0.0006 mole of HCl reacts with 0.0006 mole of sodium carbonate.

<u>Moles of sodium carbonate = 0.0006 moles</u>

5 0
3 years ago
How many moles are in 3.0x10^24 molecules of water​
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How many grams of carbon dioxide will be produced if 76.4 grams of
goldfiish [28.3K]

Answer:

Assuming that all of the oxygen is used up, 1.53×4111.53×411 or 0.556 moles of C2H3Br3 are required. Because there are only 0.286 moles of C2H3Br3 available, C2H3Br3 is the limiting reagent.

Limiting Reagent What is the limiting reagent if 76.4 grams of C2H3Br3 were reacted with 49.1 grams of O2? C2H3Br3 + 11O2 → 8CO2 + 6H2O + 6Br2 SOLUTION Using Approach 1: A. 76.4g &times; (1 mol/ 266.72 g) = 0.286 moles C2H3Br3 49.1g &times; (1 mole/ 32 g) = 1.53 moles O2 B.

Explanation:

MRK ME BRAINLIEST PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ

https://chem.libretexts.org/Bookshelves/Introductory_Chemistry/Map%3A_Introductory_Chemistry_(Tro)/08%3A_Quantities_in_Chemical_Reactions/8.04%3A_Limiting_Reactant_and_Theoretical_Yield

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3 years ago
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please brainleast

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