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1 year ago

Who was the first person to show strong empirical evidence for the existence of atoms?.

1 answer:

6 0

Given what we know, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

<h3>Who was John Dalton?</h3>

He was a renowned scientist with knowledge in many fields.
He was known to be a chemist, meteorologist, and physicist.
He proposed the atomic theory and carried out experiments to provide supporting evidence.

Therefore, given his proposal of the atomic theory and the experiments he carried out to provide evidence to support his claims, we can confirm that John Dalton was the first person to show strong empirical evidence for the existence of atoms.

To learn more about atoms visit:

brainly.com/question/13981855?referrer=searchResults

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PLEASE HELP ME I NEED HELP

Crazy boy [7]

Answer:

5 moles S x (36.02 g S/mole S) = 180.1 grams of S

Explanation:

The periodic table has mass units for every element that can be correlated with the number of atoms of that element. The relationship is known as Avogadro's Number. This number, 6.02x $10^{23}$ , is nicknamed the mole, which scientists found to be a lot more catchy, and easier to write than 6.02x $10^{23}$ . The mole is correlated to the atomic mass of that element. The atomic mass of sulfur, S, is 36.02 AMU, atomic mass units. But it can also be read as 36.02 grams/mole.

This means that 36.02 grams of S contains 1 mole (6.02x $10^{23}$ ) of S atoms.

This relationship holds for all the elements. Zinc, Zn, has an atomic mass of 65.38 AMU, so it has a "molar mass" of 65.38 grams/mole. ^5.38 grams of Zn contains 1 mole of Zn atoms.

And so on.

5.0 moles of Sulfur would therefore contain:

(5.0 moles S)*(36.02 grams/mole S) = 180.1 grams of S

Note how the units cancel to leaves just grams. The units are extremely helpful in mole calculations to insure the correct mathematical operation is done. To find the number of moles in 70 g of S, for example, we would write:

(70g S)/(36.02 grams S/mole S) = 1.94 moles of S. [Note how the units cancel to leave just moles]

4 0

2 years ago

what is the diffrence between displacement ans double displacement reactions write equations for these reactions

Lina20 [59]

Answer:

Single Displacement reaction

In a displacement reaction, a more reactive element replaces a less reactive element from a compound.

Change in colour takes place with no precipitate forms.

Metals react with the salt solution of another metal.

Examples:

2KI + Cl2 → 2KCl + I2

CuSO4 + Zn → ZnSO4 + Cu

Double displacement reaction

In a double displacement reaction, two atoms or a group of atoms switch places to form new compounds.

Precipitate is formed.

Salt solutions of two different metals react with each other.

Examples:

Na2SO4 + BaCl2 → BaSO4 + 2NaCl

2KBr + BaCl2 → 2KCl + BaBr2

Hope this helps...Please Mark as Brainliest!!

8 0

3 years ago

What is wave particle duality

yKpoI14uk [10]

Wave particle duality is the concept in quantum mechanics that every particle or quantum entity may be described as either a particle or a wave. It expresses the inability of the classical concepts "particle" or "wave" to fully describe the behaviour of quantum-scale objects.

4 0

3 years ago

What is the atomic number of beryllium

Alexandra [31]

Answer:

4 the symbol is Be

Explanation:

hope this helped

8 0

2 years ago

Read 2 more answers

Oh, no! You just spilled 85.00 mL of 1.500 M sulfuric acid on your lab bench and need to clean it up immediately! Right next to

vredina [299]

Explanation:

We will balance equation which describes the reaction between sulfuric acid and sodium bicarbonate: as follows.

$H_2SO_4(aq) + 2NaHCO_3(s) \rightarrow Na2SO_4(aq) + 2H_2O(l) + 2CO_2(g)$

Next we will calculate how many moles of $H_2SO_4$ are present in 85.00 mL of 1.500 M sulfuric acid.

As, Molarity = $\frac{\text{moles of solute}}{\text{liters of solution
}}$

1.500 M = $\frac{n}{0.08500 L
}$

n = 0.1275 mol $H_2SO_4$

Now set up and solve a stoichiometric conversion from moles of $H_2SO_4$ to grams of $NaHCO_3$ . As, the molar mass of $NaHCO_3$ is 84.01 g/mol.

$0.1275 mol H_2SO_4 \times (\frac{2 mol NaHCO_3}{1 mol H_2SO_4}) \times (\frac{84.01 g NaHCO_3}{1 mol NaHCO_3})$

= 21.42 g $NaHCO_3$

So unfortunately, 15.00 grams of sodium bicarbonate will "not" be sufficient to completely neutralize the acid. You would need an additional 6.42 grams to complete the task.

4 0

2 years ago