John Dalton believed which of the following about atoms?

Atoms are real even though they're invisible.
The atom could be divided into smaller parts.
All atoms of a single substance are identical.
Atoms of different substances differ by weight.

Answers

Answer 1

Atoms of different substances differ by weight. Option D

A) Atoms are real even though they're invisible: Dalton proposed that atoms are fundamental, indivisible particles that make up all matter. While atoms themselves cannot be observed directly, their existence and behavior can be inferred through their effects on matter.

B) The atom could be divided into smaller parts: Initially, Dalton believed that atoms were indivisible and the ultimate building blocks of matter. However, subsequent scientific discoveries, such as the discovery of subatomic particles like protons, neutrons, and electrons, revealed that atoms could be further divided into smaller components.

C) All atoms of a single substance are identical: Dalton postulated that atoms of the same element are identical in size, mass, and chemical properties. According to his atomic theory, different elements are composed of unique atoms, and atoms of the same element are identical to one another.

D) Atoms of different substances differ by weight: Dalton recognized that atoms have different masses and proposed that the differences in atomic weight account for the distinct properties of different elements. He formulated the law of multiple proportions, which states that elements combine in fixed ratios of masses to form compounds.

Option D

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Related Questions

The composition of a compound is 28.73% K, 1.48% H, 22.76% P, and 47.03% O. The molar mass of the
compound is 136.1 g/mol.
I

Answers

The compound has an empirical formula of [tex]K_2H_2P_2O_8[/tex] and a molecular formula of [tex]K_2HPO_4[/tex].

The given compound has a percent composition of K = 28.73%, H = 1.48%, P = 22.76%, and O = 47.03%. Its molar mass is 136.1 g/mol. To determine its molecular formula, we need to find its empirical formula and calculate its molecular formula from its empirical formula.The empirical formula is the smallest whole number ratio of atoms in a compound. It can be determined by converting the percent composition of the elements into their respective moles and dividing each by the smallest number of moles calculated. The moles of K, H, P, and O in 100 g of the compound are: K = 28.73 g x (1 mol/39.1 g) = 0.734 molH = 1.48 g x (1 mol/1.01 g) = 1.46 molP = 22.76 g x (1 mol/30.97 g) = 0.736 molO = 47.03 g x (1 mol/16.00 g) = 2.94 molDividing each by the smallest number of moles gives the following ratios: K = 0.734/0.734 = 1H = 1.46/0.734 = 2P = 0.736/0.734 = 1.002O = 2.94/0.734 = 4. The empirical formula of the compound is [tex]K_2H_2P_2O_8[/tex]. To calculate the molecular formula, we need to determine the factor by which the empirical formula should be multiplied to obtain the molecular formula. This can be done by comparing the molar mass of the empirical formula to the molar mass of the compound.The molar mass of [tex]K_2H_2P_2O_8[/tex] is: [tex]M(K_2H_2P_2O_8)[/tex] = (2 x 39.1 g/mol) + (2 x 1.01 g/mol) + (2 x 30.97 g/mol) + (8 x 16.00 g/mol) = 276.2 g/mol. The factor by which the empirical formula should be multiplied is: M(molecular formula)/M(empirical formula) = 136.1 g/mol/276.2 g/mol = 0.4935. The molecular formula is obtained by multiplying the empirical formula by this factor: [tex]K_2H_2P_2O_8 * 0.4935 = K_2HPO_4[/tex]. Therefore, the molecular formula of the compound is [tex]K_2HPO_4[/tex].The molecular formula of the given compound having a composition of 28.73% K, 1.48% H, 22.76% P, and 47.03% O with a molar mass of 136.1 g/mol is [tex]K_2HPO_4[/tex]. The empirical formula of the compound is [tex]K_2H_2P_2O_8[/tex]. The compound's molecular formula is calculated by determining the factor by which the empirical formula should be multiplied to obtain the molecular formula. The factor is M(molecular formula)/M(empirical formula) = 136.1 g/mol/276.2 g/mol = 0.4935. The molecular formula of the compound is obtained by multiplying the empirical formula by this factor, resulting in the molecular formula [tex]K_2HPO_4[/tex].

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The correct question would be as

The composition of a compound is 28.73% K. 1.48% H, 22.76% P, and 47.03% O. The molar mass of the compound is 136.1 g/mol. What is the Molecular Formula of the compound?

[tex]KH_2PO_4\\KH_3PO_4\\K_2H_4P_20_{12}\\K_2H_3PO_6[/tex]

Mention three significant of water in coal fired power station

Answers

Water in coal-fired power stations is used for cooling, steam generation, and pollution control, including capturing sulfur dioxide and cooling exhaust gases. Efficient water recycling helps minimize environmental impact.

Water plays a critical role in coal-fired power stations. The power stations need large quantities of water for a variety of purposes. Water is primarily used to cool the power plant, maintain a safe temperature in the boilers, and also to generate steam. In this context, this answer will discuss three significant uses of water in coal-fired power stations. Significant uses of water in coal-fired power stations1. Cooling: Power stations require water for cooling purposes. When water is used for cooling, it absorbs the heat produced by the combustion process. Cooling towers are responsible for releasing the heated water, which is then reused.2. Steam generation: Water is required to generate steam, which is used to rotate turbines and generate electricity. The water used to generate steam must be treated to prevent the accumulation of harmful minerals, which can damage the power plant.3. Pollution control: Water is utilized to reduce air pollution. Flue gas desulfurization systems utilize water to capture sulfur dioxide from power plants. Water is also used to cool exhaust gases that are produced during combustion.In conclusion, the three significant uses of water in coal-fired power stations include cooling, steam generation, and pollution control. These processes require large amounts of water, which is why coal-fired power stations are often located near water sources. By recycling water, power stations can conserve water and minimize their environmental impact.

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Calculate the mass of wire that reacted to silver nitrate solution Mass being 1.52 of copper before reaction

Answers

The mass of wire that reacted to silver nitrate solution is 5.15 grams.

To calculate the mass of the wire that reacted with silver nitrate solution, we need to consider the stoichiometry of the reaction. The reaction between copper and silver nitrate can be represented by the following equation:

Cu + 2AgNO3 → Cu(NO3)2 + 2Ag

According to the equation, one mole of copper reacts with two moles of silver nitrate to form one mole of copper(II) nitrate and two moles of silver.

Given that the mass of copper before the reaction is 1.52 grams, we can calculate the molar mass of copper using its atomic mass, which is 63.55 grams/mol.

1.52 g of copper is equal to 1.52 g / 63.55 g/mol = 0.0239 moles of copper.

Since the reaction stoichiometry is 1:2 between copper and silver, the moles of copper reacting would be equal to half of the moles of silver formed.

Therefore, the moles of silver formed would be 0.0239 moles x 2 = 0.0478 moles.

To find the mass of silver, we multiply the moles of silver by its molar mass, which is 107.87 grams/mol:

Mass of silver = 0.0478 moles x 107.87 g/mol = 5.15 grams.

Hence, the mass of wire that reacted to silver nitrate solution is 5.15 grams.

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I think it is the question:

A copper wire with a mass of 1.52 grams reacted with silver nitrate solution. If the balanced chemical equation and the molar ratio between copper and silver nitrate are provided, how can you determine the mass of wire that reacted?

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