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.
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A. Dalton's theory that atoms could not be divided was incorrect
<em>Calculate the pH of the following substances formed during a volcanic eruption:
</em>
<em>• Acid rain if the [H +] is 1.9 x 10-5
</em>
<em>• Sulfurous acid if [H +] = 0.10
</em>
<em>• Nitric acid if [H +] = 0.11</em>
<em />
<h3>Further explanation </h3>
pH is the degree of acidity of a solution that depends on the concentration of H⁺ ions. The greater the value the more acidic the solution and the smaller the pH.
pH = - log [H⁺]
![\tt pH=-log[1.9\times 10^{-5}]\\\\pH=5-log1.9\\\\pH=4.72](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B1.9%5Ctimes%2010%5E%7B-5%7D%5D%5C%5C%5C%5CpH%3D5-log1.9%5C%5C%5C%5CpH%3D4.72)
![\tt pH=-log[10^{-1}]\\\\pH=1](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B10%5E%7B-1%7D%5D%5C%5C%5C%5CpH%3D1)
![\tt pH=-log[11\times 10^{-2}]\\\\pH=2-log~11=0.959](https://tex.z-dn.net/?f=%5Ctt%20pH%3D-log%5B11%5Ctimes%2010%5E%7B-2%7D%5D%5C%5C%5C%5CpH%3D2-log~11%3D0.959)
Answer: 7
Explanation:
Before a number but after a decimal. The zeros at the end would usually mean that it doesn't count but since the numbers are before the zeros and after a decimal it's 7 sig figs
Answer:
The chemist would require to use 43.43 grams.
Explanation:
In order to solve this problem we need to know<u> how much do 0.550 moles of selenium weigh</u>. To do that we use selenium's<em> molar mass </em>and multiply it by the given number of moles:
- 0.550 mol * 78.96 g/mol = 43.43 g
The chemist would require to use 43.43 grams.
