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Sphinxa [80]
3 years ago
15

Which of the following measurements is expressed to three significant figures?

Chemistry
2 answers:
Norma-Jean [14]3 years ago
6 0

Answer:

D

Explanation:

it goes past the decimal three places

LenKa [72]3 years ago
4 0

Answer:

C

Explanation:

A is  1 sig fig

B is 4 sig fig

C is 3 sig fig

D is 2 sig fig

Decimal places do not necessarily determine sig figs.

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to find the density of stopper I weighted it and found its mass to 4.8g. After that I filled a graduated cylinder with 32.1mL of
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Answer:

0.68g/ml

Explanation:

The density of an object is its mass per unit volume. It is calculated using the formula

Density = mass / volume

Mass of stopper weighed = 4.8g

The volume of stopper can be got by subtracting the (volume of water) from the (volume of water+stopper) i.e.

= 39.2ml - 32.1ml

= 7.1ml

Volume of stopper = 7.1ml

Density of stopper= 4.8/7.1

Density= 0.676056

Therefore, the density of the stopper is 0.68g/ml

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The atomic masses of the elements _____ as you move left to right on the periodic table
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Consider the unbalanced equation for aluminum and oxygen producing aluminum oxide: Al + O₂ → Al₂O₃. Which of the following shows
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3 years ago
A 27.9 mL sample of 0.289 M dimethylamine, (CH3)2NH, is titrated with 0.286 M hydrobromic acid.
sesenic [268]

Answer:

(1) Before the addition of any HBr, the pH is 12.02

(2) After adding 12.0 mL of HBr, the pH is 10.86

(3) At the titration midpoint, the pH is 10.73

(4) At the equivalence point, the pH is 5.79

(5) After adding 45.1 mL of HBr, the pH is 1.18

Explanation:

First of all, we have a weak base:

  • 0 mL of HBr is added

(CH₃)₂NH  + H₂O  ⇄  (CH₃)₂NH₂⁺  +  OH⁻            Kb = 5.4×10⁻⁴

0.289 - x                             x                x

Kb = x² / 0.289-x

Kb . 0.289 - Kbx - x²

1.56×10⁻⁴ - 5.4×10⁻⁴x - x²

After the quadratic equation is solved x = 0.01222 → [OH⁻]

- log  [OH⁻] = pOH → 1.91

pH = 12.02   (14 - pOH)

  • After adding 12 mL of HBr

We determine the mmoles of H⁺, we add:

0.286 M . 12 mL = 3.432 mmol

We determine the mmoles of base⁻, we have

27.9 mL . 0.289 M = 8.0631 mmol

When the base, react to the protons, we have the protonated base plus water (neutralization reaction)

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       3.432 mm                 -

4.6311 mm                                  3.432 mm

We substract to the dimethylamine mmoles, the protons which are the same amount of protonated base.

[(CH₃)₂NH] → 4.6311 mm / Total volume (27.9 mL + 12 mL) = 0.116 M

[(CH₃)₂NH₂⁺] → 3.432 mm / 39.9 mL = 0.0860 M

We have just made a buffer.

pH = pKa + log (CH₃)₂NH  / (CH₃)₂NH₂⁺

pH = 10.73 + log (0.116/0.0860) = 10.86

  • Equivalence point

mmoles of base = mmoles of acid

Let's find out the volume

0.289 M . 27.9 mL = 0.286 M . volume

volume in Eq. point = 28.2 mL

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm       8.0631mm               -

                                                8.0631 mm

We do not have base and protons, we only have the conjugate acid

We calculate the new concentration:

mmoles of conjugated acid / Total volume (initial + eq. point)

[(CH₃)₂NH₂⁺] = 8.0631 mm /(27.9 mL + 28.2 mL)  = 0.144 M

(CH₃)₂NH₂⁺   +  H₂O   ⇄   (CH₃)₂NH  +  H₃O⁻       Ka = 1.85×10⁻¹¹

 0.144 - x                                  x               x

[H₃O⁺] = √ (Ka . 0.144) →  1.63×10⁻⁶ M  

pH = - log [H₃O⁺] = 5.79

  • Titration midpoint (28.2 mL/2)

This is the point where we add, the half of acid. (14.1 mL)

This is still a buffer area.

mmoles of H₃O⁺ = 4.0326 mmol (0.286M . 14.1mL)

mmoles of base = 8.0631 mmol - 4.0326 mmol

[(CH₃)₂NH] = 4.0305 mm / (27.9 mL + 14.1 mL) = 0.096 M

[(CH₃)₂NH₂⁺] = 4.0326 mm (27.9 mL + 14.1 mL) = 0.096 M

pH = pKa + log (0.096M / 0.096 M)

pH = 10.73 + log 1 =  10.73

Both concentrations are the same, so pH = pKa. This is the  maximum buffering capacity.

  • When we add 45.1 mL of HBr

mmoles of acid = 45.1 mL . 0.286 M = 12.8986 mmol

mmoles of base = 8.0631 mmoles

This is an excess of H⁺, so, the new [H⁺] = 12.8986 - 8.0631 / Total vol.

(CH₃)₂NH     +      H₃O⁺        ⇄  (CH₃)₂NH₂⁺  +  H₂O

8.0631 mm     12.8986 mm             -

       -               4.8355 mm                        

[H₃O⁺] = 4.8355 mm / (27.9 ml + 45.1 ml)

[H₃O⁺] = 4.8355 mm / 73 mL → 0.0662 M

- log [H₃O⁺] = pH

- log 0.0662 = 1.18 → pH

7 0
3 years ago
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