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Sergio [31]
1 year ago
10

An empty Erlenmeyer flask weighs 241.3 g. When filled with water (d = 1.00 g/cm³), the flask and its contents weigh 489.1 g.

Chemistry
1 answer:
Viefleur [7K]1 year ago
4 0

The answer is- Mass of flask = 608.04 g

Density (d) is defined as mass of substance divided by its volume. Thus, if mass and density are known, then Volume can be determined.

What is the formula of  mass in terms of density?

  • Let mass of substance be 'm' and volume be 'V'. Thus, as per the definition of density, it is expressed as-

d = \frac{m}{V}

Thus, mass can be expressed as-

m = d * V

  • Now, mass of empty flask = 241.3 g and the mass of  (flask + Water) = 489.1 g.

Thus, mass of water = 489.1\ g- 241.3\ g = 247.8\ g.

  • Thus, mass of water = 247.8 g and density of water =1.00\ g/cm^3. Its volume is calculated as-

V = \frac{247.8\ g}{1.00\ g/cm^3}=247.8\ cm^3

  • Volume of flask = 247.8 cm^3 .
  • Then when this flask is filled with chloroform (d =1.48 g/cm^3), the mass of chloroform is-

m = d * V = (1.48\ g/cm^3) * (247.8\ cm^3)= 366.74\ g

  • Hence, the mass of flask becomes =241.3\ g + 366.74\ g = 608.04\ g

To learn more about Density and mass, visit:

brainly.com/question/952755

#SPJ4

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Answer:

\large \boxed{1.77 \times 10^{-5}\text{ mol/L}}

Explanation:

Assume that you have mixed 135 mL of 0.0147 mol·L⁻¹ NiCl₂ with 190 mL of 0.250 mol·L⁻¹ NH₃.

1. Moles of Ni²⁺

n = \text{135 mL} \times \dfrac{\text{0.0147 mmol}}{\text{1 mL}} = \text{1.984 mmol}

2. Moles of NH₃

n = \text{190 mL} \times \dfrac{\text{0.250 mmol}}{\text{1 mL}} = \text{47.50 mmol}

3. Initial concentrations after mixing

(a) Total volume

V = 135 mL + 190 mL = 325 mL

(b) [Ni²⁺]

c = \dfrac{\text{1.984 mmol}}{\text{325 mL}} = 6.106 \times 10^{-3}\text{ mol/L}

(c) [NH₃]

c = \dfrac{\text{47.50 mmol}}{\text{325 mL}} = \text{0.1462 mol/L}

3. Equilibrium concentration of Ni²⁺

The reaction will reach the same equilibrium whether it approaches from the right or left.

Assume the reaction goes to completion.

                        Ni²⁺             +             6NH₃       ⇌       Ni(NH₃)₆²⁺

I/mol·L⁻¹:    6.106×10⁻³                     0.1462                       0

C/mol·L⁻¹:  -6.106×10⁻³         0.1462-6×6.106×10⁻³             0

E/mol·L⁻¹:           0                              0.1095                6.106×10⁻³

Then we approach equilibrium from the right.

                            Ni²⁺   +   6NH₃       ⇌       Ni(NH₃)₆²⁺

I/mol·L⁻¹:              0           0.1095                6.106×10⁻³

C/mol·L⁻¹:            +x            +6x                           -x

E/mol·L⁻¹:             x         0.1095+6x            6.106×10⁻³-x

K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}

Kf is large, so x ≪ 6.106×10⁻³. Then

K_{\text{f}} = \dfrac{\text{[Ni(NH$_{3}$)$_{6}^{2+}$]}}{\text{[Ni$^{2+}$]}\text{[NH$_{3}$]}^{6}} = 2.0 \times 10^{8}\\\\\dfrac{6.106 \times 10^{-3}}{x\times 0.1095^{6}} = 2.0 \times 10^{8}\\\\6.106 \times 10^{-3} = 2.0 \times 10^{8}\times 0.1095^{6}x= 345.1x\\x= \dfrac{6.106 \times 10^{-3}}{345.1} = 1.77 \times 10^{-5}\\\\\text{The concentration of Ni$^{2+}$ at equilibrium is $\large \boxed{\mathbf{1.77 \times 10^{-5}}\textbf{ mol/L}}$}

 

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