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3 years ago

Which states of matter have particles that move independently of one another with very little attraction?

2 answers:

maks197457 [2]3 years ago

8 0

Gas -- move very independently as compared to a liquid, for instance.

Plasma -- related to a gas although it has positively charged ions and negatively charged electrons.

pychu [463]3 years ago

5 0

what was the point of deleting my comment... that was irrelevant.

You might be interested in

Zn + 2 HCl → ZnCl2 + H2

MrRissso [65]

Answer:

10.1g of H₂ are produced

Explanation:

To solve this question we need, first, to convert the mass of each reactant to moles and, using the chemical reaction, find limiting reactant. With limiting reactant we can find the moles of H2 and its mass:

Moles Zn -Molar mass: 65.38g/mol-:

307g * (1mol / 65.38g) = 4.696 moles

Moles HCl -Molar mass: 36.46g/mol-:

381g HCl * (1mol / 36.46g) = 10.45 moles

For a complete reaction of 10.45 moles of HCl are required:

10.45 moles HCl * (1mol Zn / 2mol HCl) = 5.22 moles Zn

As there are 4.696 moles of Zn, Zn is the limiting reactant

The moles of H₂ produced = Moles of Zn added = 4.696 moles. The mass is-Molar mass H₂ = 2.16g/mol-:

4.696 moles * (2.16g / mol) =

<h3>10.1g of H₂ are produced</h3>

7 0

3 years ago

CaCO3(s) ∆→CaO(s) + CO2(g).

sveta [45]

Answer:

74.9%.

Explanation:

Relative atomic mass data from a modern periodic table:

Ca: 40.078;
C: 12.011;
O: 15.999.

What's the theoretical yield of this reaction?

In other words, what's the mass of the CO₂ that should come out of heating 40.1 grams of CaCO₃?

Molar mass of CaCO₃:

$M(\text{CaCO}_3) = 40.078 + 12.011 + 3 \times 15.999 = 100.086\;\text{g}\cdot\text{mol}^{-1}$ .

Number of moles of CaCO₃ available:

$\displaystyle n(\text{CaCO}_3) =\frac{m}{M} = \frac{40.1}{100.086} = 0.400655\;\text{mol}$ .

Look at the chemical equation. The coefficient in front of both CaCO₃ and CO₂ is one. Decomposing every mole of CaCO₃ should produce one mole of CO₂.

$n(\text{CO}2) = n(\text{CaCO}_3)= 0.400655\;\text{mol}$ .

Molar mass of CO₂:

$M(\text{CO}_2) = 12.011 + 2\times 15.999 = 44.009\;\text{g}\cdot\text{mol}^{-1}$ .

Mass of the 0.400655 moles of $\text{CO}_2$ expected for the 40.1 grams of CaCO₃:

$m(\text{CO}_2) = n\cdot M = 0.400655 \times 44.009 = 17.632\;\text{g}$ .

What's the percentage yield of this reaction?

$\displaystyle \textbf{Percentage}\text{ Yield} = \frac{\textbf{Actual}\text{ Yield}}{\textbf{Theoretical}\text{ Yield}}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} = \frac{13.2}{17.632}\times 100\%\\\phantom{\textbf{Percentage}\text{ Yield}} =74.9\%$ .

7 0

3 years ago

A 35.161 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion analysis ap

xxMikexx [17]

Answer:

The empirical formulae is C6H12S02

Explanation:

1. First we need to obtain the mass of each element in the sample and compound formed

Carbon = (62.637 mg * 12.011 g/mol / 44.009 g/mol) = 17.094 mg of Carbon

Hydrogen = ( 25.641 mg * (2 *1..008 g/mol) / 18.015 g/mol) = 2.869 mg of Hydrogen

Sulphur = (13.54 mg * 32.066 g/mol / 64.066 g/mol) = 6.777 mg of Sulphur

2. Next is to determine the percentage composition. Here we divide the respective mass by the mass of the sample

Carbon = 17.094 / 35.161 * 100 = 48.62 %

Hydrogen = 2.869/ 35.161 *100 = 8.16 %

Sulphur = 6.777/ 31.321 *100 = 21.64 %

Oxygen = (100 - (48.62 + 8.16 + 21.64)) = 21.58 %

3. Next is to divide the mass assuming there are 100 mg by the respective atomic masses to obtain the number of moles

Carbon = 48.62 / 12.011 = 4.048 mol

Hydrogen = 8.16 / 1.008 = 8.095 mol

Sulphur = 21.64 / 32.066 = 0.675 mol

Oxygen = 21.58 / 16.000 = 1.348 mol

Next is to divide by the smallest value

Carbon = 4.048/ 0.675 =5.997 = 6

Hydrogen = 8.095 / 0.675 =11.993 =12

Sulphur = 0.675/ 0.675 = 1

Oxygen = 1.348 / 0.675 = 1.997 = 2

So therefore the empirical formulae of the sample is C6H12SO2

7 0

3 years ago

Help please ? Thank you

olasank [31]

Answer:

40J

Explanation:

I think about that why I don't know what is the answer

3 0

3 years ago

Calculate the molar solubility of silver(i) bromate with ksp = 5. 5×10-5. also, convert the molar solubility to the solubility.

Sindrei [870]

The molar solubility is 7.4× $10^{-3}$ M and the solubility is 7.4× $10^{-3}$ g/L .

Calculation ,

The dissociation of silver bromide is given as ,

$AgBr$ → $Ag ^{+}$ + $Br^{-}$

- S S

Ksp = [ $Ag ^{+}$ ] [ $Br^{-}$ ] = [S] [ S ] = $S^{2}$

S = √ Ksp = √ 5. 5× $10^{-5}$ = 7.4× $10^{-3}$

The solubility =7.4× $10^{-3}$ g/L

The molar solubility is the solubility of one mole of the substance.

Since , one mole of $AgBr$ is dissociates and form one mole of each $Ag ^{+}$ and $Br^{-}$ ion . So, solubility is equal to molar solubility but unit is different.

Molar solubility = 7.4× $10^{-3}$ mol/L = 7.4× $10^{-3}$ M

To learn more about molar solubility ,

brainly.com/question/16243859

#SPJ4

7 0

1 year ago