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garik1379 [7]
2 years ago
5

How do solve for density?

Chemistry
1 answer:
Pachacha [2.7K]2 years ago
7 0
Density(D) is defined as Mass(M) divided by Volume(V).

The formula for Density is:

D = M / V.

Another way to remember the formula for Density is to remember "Mass per unit of volume".

I hope this helps!
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What is one way to test whether an unknown solution is acidic or basic
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A solution's pH will be a number between 0 and 14. A solution with a pH of 7 is classified as neutral. If the pH is lower than 7, the solution is acidic. When pH is higher than 7, the solution is basic. These numbers describe the concentration of hydrogen ions in the solution and increase on a negative logarithmic scale. For example, If Solution A has a pH of 3 and Solution B has a pH of 1, then Solution B has 100 times as many hydrogen ions than A and is therefore 100 times more acidic.

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I need help understanding polyatomic ions
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3 0
3 years ago
You wish to measure the iron content of the well water on the new property you are about to buy. You prepare a reference standar
djverab [1.8K]

1.04 ⨯ 10^{-4} M

<h3>Explanation</h3>

<em>A</em> = <em>ε</em> \cdot l \cdot c by the Beer-Lambert law, where

  • <em>A</em> the absorbance,
  • l the path length,
  • <em>ε</em> the molar absorptivity of the solute, and
  • c concentration of the solution.

<em>A</em> and <em>ε </em>are the same for both solutions. Therefore, l \cdot c is constant; l is inversely proportional to c. The 100 mL sample would have a concentration 1/4.78 times that of the 45.0 mL reference.

The 13.0 mL standard solution has a concentration of 5.17 ⨯ 10^{-4} M. Diluting it to 45.0 mL results in a concentration of 5.17 \times 10^{-4} \times \frac{13.0}{45.0} = 1.494 M.

c is inversely related to l for the two solutions. As a result, c₂ = c_1 \cdot \frac{l_1}{l_2} = 1.494 \times 10^{-4} \times \frac{1}{4.78} = 3.126 M.

The 30.0 mL sample has to be diluted by 30.0 / 100.0 times to produce the 100.0 mL solution being tested. The 100.0 mL solution has a concentration of 3.126 M. Therefore, the 30.0 mL solution has a concentration of 3.126 \times \frac{100.0}{30.0} = 1.04 ⨯ 10^{-4} M.

6 0
3 years ago
A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat
Jet001 [13]
Triprotic acid is a class of Arrhenius acids that are capable of donating three protons per molecule when dissociating in aqueous solutions.  So the chemical reaction as described in the question, at the third equivalence point, can be show as: H3R + 3NaOH ⇒ Na3R + 3H2O, where R is the counter ion of the triprotic acid. Therefore, the ratio between the reacted acid and base at the third equivalence point is 1:3. 
The moles of NaOH is 0.106M*0.0352L = 0.003731 mole.  So the moles of H3R is 0.003731mole/3=0.001244mole.
The molar mass of the acid can be calculated: 0.307g/0.001244mole=247 g/mol.
6 0
3 years ago
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