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Firlakuza [10]
2 years ago
11

Write the advantages of molecular formula​

Chemistry
1 answer:
dusya [7]2 years ago
8 0

Answer:

Its main advantage is <em>they information fits on one line of text</em> <u>(thus works well when using the formula in paragraphs)</u>. Disadvantages are <em>they can be confusing for larger molecules</em>

<em />

hope this helps :3

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The many rows of sharp, jagged teeth on some sharks is an example of an
RSB [31]
Maybe venus fly trap
6 0
4 years ago
Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlori
Helga [31]

Answer:

pH = 6.82

Explanation:

To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:

  • pH = pKa + log\frac{[NaOCl]}{[HOCl]}

We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:

  • pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54
  • [HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M
  • pH = 7.54 + log \frac{4}{4}
  • pH = 7.54

By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:

  • pH = 7.54 + log\frac{[NaOCl-HCl]}{[HOCl+HCl]}
  • pH = 7.54 + log \frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}
  • pH = 6.82
6 0
3 years ago
the diagram below represents two beakers, each containing an ice cube and clear liquid. in beaker A the ice cube floats, and in
Mama L [17]

Answer:

The density of the liquid in beaker B is less than the that of ice.

Explanation:

Ice will float if its mass is less than the mass of the liquid it displaces.

For example, the density of ice is less than that of water.

A 10 cm³ cube of ice has a mass of about 9 g, while the mass of 10 cm³ of water is 10 g. Thus, 9 g of ice displaces 10 g of water.

The denser water displaces the lighter ice and the ice floats to the top.

If the density of the liquid is <em>less than</em> that of water, say, 8 g/cm³, the ice will displace only 8 g of the liquid. The ice will sink.

4 0
4 years ago
A 20.00-ml sample of 0.3000 m hbr is titrated with 0.15 m naoh. what is the ph of the solution after 40.3 ml of naoh have been a
Zielflug [23.3K]

Molarity = no.of moles/vol.of the solution

Then, no.of moles of NaOH= 0.15M*40.3ml

= 6.04510x^{-3}  moles

no.of moles of HCl= 0.30M×20ml= 6.010^-3 moles

The reaction is as bellows:

HCl+NaOH= NaCl+H2O

1 moleHCl reacts with 1 mole NaOH

Therefore,6.0 × 10^- .. 6.0 × 10 ^-3 moles of NaOH

Then, unreacted NaOH= (6.045×10^-3–6.0×10^-3) moles

= 4.5×10^-5 moles

Volume of the solution = (20+40.3)ml= 60.3ml= 6.03×10–2 L

Now, molar concentration of NaOH= 4.5×10^-5 moles/6.03×10^-2L

= 7.46×10–4 moles/L

NaOH completely dissociates in solution.

Therefore,[NaOH]=[[OH-]=7.46×10^-4 moles/L

pOH= -log[OH-]= -log(7.46×10^-4)= 3.12

Again,pH+pOH=14

or,pH=14-pOH

= 14–3.12= 10.88

<h3> </h3><h3>If sodium hydroxide were added to a solution of strong acid, what would happen to the pH of the solution?</h3>

Depending on how much NaOH is added, yes. The solution has a pH of 7 if the amount of additional NaOH is exactly equal to the amount of acid.

Since NaOH is an excess reagent, the solution will become basic if more equivalents of base are added than equivalents of acid, and the pH of the solution will rise above 7.

The solution will be acidic, with a pH lower than 7, if the number of equivalents of acid exceeds that of the NaOH that has been added.

To learn more about Sodium hydroxide,visit:

brainly.com/question/20371039

#SPJ4

8 0
2 years ago
If the reaction consumes methane gas ( CH4 ) at a rate of 2.08 M/s, what is the rate of formation of H2 ? the balanced equation
alexira [117]

Answer:

4.16M/s

Explanation:

Based on the reaction:

CH₄ + N₂Cl₄ ⇄ CCl₄ + N₂ + 2H₂

1 mole of methane, CH₄, produce 2 moles of H₂.

That means whereas 1 mole of methane is consumed, 2 moles of H₂ are formed

Having this in mind, if you are consuming methane at a rate of 2.08M/s, the rate of formation of hydrogen must be twice this rate, because there are produced twice moles of H₂.

Thus, rate of formation of H₂ is:

2.08M/s ₓ 2 =

4.16M/s

6 0
4 years ago
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