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nexus9112 [7]
3 years ago
10

Which is a component of John Dalton’s atomic theory?

Chemistry
2 answers:
saul85 [17]3 years ago
8 0
C, he stated that if you broke down atoms further, they would no longer maintain its characteristics
krek1111 [17]3 years ago
3 0

Answer;

C. An atom is a small particle that cannot be broken down

Explanation;

Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

Dalton’s atomic theory also stated that all compounds were composed of combinations of these atoms in defined ratios.

Dalton also postulated that chemical reactions resulted in the rearrangement of the reacting atoms.

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When an acid reacts with a strong base which product always forms
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Answer:طيزي

Explanation:

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3 years ago
Which statements true about the total mass of the reactants during a chemical change? (5 points)
Lapatulllka [165]
It is greater than the total mass
5 0
3 years ago
Oxalic Acid, a compound found in plants and vegetables such as rhubarb, has a mass percent composition of 26.7% C, 2.24% H, and
blondinia [14]

Answer:

HCO₂

Explanation:

From the information given:

The mass of the elements are:

Carbon C = 26.7 g;     Hydrogen H = 2.24 g     Oxygen O = 71.1 g

To determine the empirical formula;

First thing is to find the numbers of moles of each atom.

For Carbon:

=26.7 \ g\times \dfrac{1 \ mol }{12.01 \ g} \\ \\ =2.22 \ mol \ of \ Carbon

For Hydrogen:

=2.24 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =2.22 \ mol \ of \ Hydrogen

For Oxygen:

=71.1 \ g\times \dfrac{1 \ mol }{1.008 \ g} \\ \\ =4.44 \ mol \ of \ oxygen

Now; we use the smallest no of moles to divide the respective moles from above.

For carbon:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ carbon

For Hydrogen:

\dfrac{2.22 \ mol \ of \ carbon}{2.22} =1 \ mol \ of \ hydrogen

For Oxygen:

\dfrac{4.44 \ mol \ of \ Oxygen}{2.22} =2 \ mol \ of \ oxygen

Thus, the empirical formula is HCO₂

4 0
3 years ago
Calculate the whole-number ratio of staggered to eclipsed conformers that are present at room temperature. Use the Actual G˚ bar
natka813 [3]

Answer:

The ratio of staggered to eclipsed conformers is 134

Explanation:

It is possible to determine the ratio of staggered to eclipsed conformers of a reactant, using the equilibrium:

Staggered ⇄ Eclipsed

Keq = [Eclipsed] / [Staggered]

That means Keq is equal to the ratio we need to find:

Using:

G˚= -RTln Keq

<em>Where G° = -12133.6J/mol</em>

<em>R is gas constant: 8.314J/molK</em>

<em>T is absolute temperature: 298K</em>

<em />

-12133.6J/mol= -8.314J/molK*298K ln Keq

4.8974 = ln Keq

134 = Keq = [Eclipsed] / [Staggered]

<h3>The ratio of staggered to eclipsed conformers is 134</h3>
3 0
3 years ago
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