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2 years ago

john dalton's original atomic theory contained the following key ideas. which part(s) of these ideas was/were incorrect?

1 answer:

7 0

John Dalton's original atomic theory contained the following key ideas and the incorrect one is that elements are made of tiny indivisible particles called atoms and is denoted as option A.

This is defined as the smallest unit of matter which forms a chemical element and Dalton proposed that it was indivisible which was later proved wrong.

It was later discovered that atom is made up of sub atomic particles such as proton, electron and neutron. This was therefore the reason why option A was chosen as the most appropriate choice.

Read more about Atom here brainly.com/question/6258301

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The options include:

A. elements are made of tiny indivisible particles called atoms

B. Atoms are unchanged in chemical reaction

C. Atoms can join together in whole number ratios to form compounds.

D. The atoms of each element are unique

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Calcula la masa atómica del Hierro y las partículas subatómicas de cada uno de sus isótopos. Fe-54 (5.82%), Fe-56 (91.66%), Fe-5

tino4ka555 [31]

Answer:

La masa atómica del hierro es 55.847 gramos por mol.

Explanation:

Las masas molares de $Fe_{54}$ , $Fe_{56}$ , $Fe_{57}$ y $Fe_{58}$ son 53.940 gramos por mol, 55.935 gramos por mol, 56.935 gramos por mol y 57.933 gramos por mol, respectivamente. La masa atómica del hierro se determina mediante el siguiente promedio ponderado:

$M_{Fe} = \frac{5.82}{100}\times \left(53.940\,\frac{g}{mol} \right)+\frac{91.66}{100}\times (55.935\,\frac{g}{mol})+\frac{2.19}{100}\times \left(56.935\,\frac{g}{mol} \right)+\frac{0.33}{100}\times \left(57.933\,\frac{g}{mol} \right)$

$M_{Fe} = 55.847\,\frac{g}{mol}$

La masa atómica del hierro es 55.847 gramos por mol.

8 0

3 years ago

PLEASE ANSWER I WILL GIVE YOU BRAINIEST!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

blondinia [14]

Answer:

Erosion

Explanation:

Erosion is not a cause of metamorphism

The causes are mostly temperature, pressure, heat from the metamorphic rock

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3 years ago

Read 2 more answers

The pH of a solution prepared by mixing 40.00 mL of 0.10 M NH3 with 50.00 mL of 0.10 M NH4Cl and 30mL of 0.05 M H2SO4 is 5.17. A

liq [111]

Answer:

Following are the answer to this question:

Explanation:

The value of pH solution is =5.17 So, the p^{OH}:

$p^{OH}$ =14-56.17

=8.823

The volume of the $NH_{3}$ = 40.00 ml

convert into the liter= 0.040L

The value of the concentrated $NH_{3}$ =0.10 M

The volume of the $NH_{4}Cl$ = 50.00 ml

convert into the liter= 0.050L

The value of concentrated $NH_{4}Cl$ = 0.10 M

The volume of the $H_{2}So_{4}$ = 30 ml

convert into the liter= 0.030L

The value of concentrated $H_2So_4$ =0.05 M

Calculating total volume=(0.40+0.050+0.030)

=0.120 L

calculating the new concentrated value of $NH_3$ = $\frac{0.10\times 0.040}{0.120}= 0.33 \ M$

calculating the new concentrated value of $NH_4Cl$ = $\frac{0.050\times 0.10}{0.120}= 0.04166 \ M$ calculating the new concentrated value of $H_2So_4= \frac{0.030\times 0.05}{0.120}= 0.0125 \ M$ when 1 mol $H_2So_4$ produced 2 mols $H^{+}$ so, 0.0125 in $H_2So_4$ produced:

$=4 \times (2 \times 0.0125) \ mol H^{+}\\\\= 0.025 mol H^{+}$

create the ICE table:

$NH_3 \ \ \ \ \ \ \ \ + H^{+} \ \ \ \ \ \ \longrightarrow NH_4^{+}$

I (m) 0.033(m) 0.025 0.04166

C -0.025 -0.025 + 0.025

E 8.3\times 10^{-3} 0 0.0667

now calculating pH:

when ph= 8.83:

$P^{H}= p^{kb}|+ \log\frac{[NH_4^{+}]}{[NH_3]}\\\\8.83=p^{kb}+\log\frac{0.0667}{8.3 \times 10^{-3}}\\\\p^{kb}=8.83-0.9069\\\\ \ \ \ =7.7231 \\\\\ The P^{kb} \ for \ NH_3 \ is =7.7231\\\\\ The P^{kb} \ for N^{+}H_4=14-7.7231\\\\\ \ \ \ \ \ =6.2769$

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gladu [14]

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alexira [117]

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