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VARVARA [1.3K]
3 years ago
7

10 points

Chemistry
1 answer:
Romashka-Z-Leto [24]3 years ago
3 0

Answer:

b

Explanation:

i justt took test

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What even is this???
Snowcat [4.5K]

Answer:

a. One particle

Explanation:

b. Two particle

6 0
3 years ago
El agua salubre es aquella que tiene más sales disueltas que el agua dulce. En el análisis de una muestra de aguas se encontró q
Slav-nsk [51]

Answer:

0.39 % m/m; 0.42 % m/v; 0.18 % v/v; 4200 ppm; 0.044 mol·L⁻¹; 0.041 mol/kg;

0.089 equiv/L; 0.000 74; 0.999 26

Explanation

Data:

Mass  of MgCl₂       =       3.8   g

Volume of solution =  900      mL

Density of solution =       1.09 g/mL

Density of MgCl₂    =      2.32 g/cm³

Calculations

1. Percent m/m

\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times 100 \, \%\\\\\text{Total mass} = \text{900 mL} \times \dfrac{\text{1.09 g}}{\text{1 mL}} = \text{981 g}\\\\\text{Mass percent} = \dfrac{\text{3.8 g}}{\text{981 g}} \times 100 \,\% = \textbf{0.39 \% m/m}

2. Percent m/v

\text{Mass-by-volume percent} = \dfrac{\text{Mass of component}}{\text{Total volume}} \times 100 \, \%\\\\\text{ Mass-by-volume percent } = \dfrac{\text{3.8 g}}{\text{900 mL}} \times 100 \,\% = \textbf{0.42 \% m/v}

3. Percent v/v

\text{Volume percent} = \dfrac{\text{Volume of component}}{\text{Total volume}} \times 100 \, \%\\\\\text{Volume of MgCl}_{2} = \text{3.8 g} \times \dfrac{\text{1 mL}}{\text{2.32 g}} = \text{1.64 mL}\\\\\text{ Mass-by-volume percent } = \dfrac{\text{1.64 mL}}{\text{900 mL}} \times 100 \,\% = \textbf{0.18 \% v/v}

4. Parts per million

\text{Ppm} = \dfrac{\text{milligrams of solute}}{\text{litres of solution}} = \dfrac{\text{3800 mg}}{\text{0.900 L}} = \textbf{ 4200 ppm}

5. Molar concentration

\text{Molar concentration} = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\\text{Moles of MgCl}_{2} = \text{3.8 g} \times \dfrac{\text{1 mol}}{\text{95.21 g}} = \text{0.040 mol}\\\\\text{Molar concentration} = \dfrac{\text{0.040 mol}}{\text{0.900 L}} = \textbf{0.044 mol/L}

6. Molal concentration

\text{Molal concentration} = \dfrac{\text{moles of solute}}{\text{kilograms of solvent}}

Mass of water = Mass of solution - mass of solute = 981 g - 3.8 g = 977 g = 0.977 kg

\text{Molar concentration} = \dfrac{\text{0.040 mol}}{\text{0.977 kg}} = \textbf{0.041 mol/kg}

7. Normality

The normality is the number of equivalents per litre of solution.

The normality of an ion equals the molar concentration times the charge on the ion.

Thus, the normality of MgCl₂ is twice the molar concentration.

Normality = 2 × 0.044 mol·L⁻¹ = 0.089 equiv·L⁻¹

8. Mole fraction of solute

\chi_{\text{solute}} = \dfrac{n_{\text{solute}}}{n_{\text{total}}}

Moles of MgCl₂ = 0.040 mol

\text{Moles of water} = \text{977 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{54.23 mol}

Total moles = n₂ + n₁ = 0.040 mol + 54.23 mol = 54.27 mol

\chi_{2} = \dfrac{\text{0.040 mol}}{\text{54.23 mol}} = \mathbf{0.00074}

9. Mole fraction of solvent

χ₁ = 1 - χ₂ = 1 - 0.000 74 = 0.999 26

4 0
3 years ago
Suppose you want to prepare a buffer with a pH of 4.59 using formic acid. What ratio of [sodium formate]/[formic acid) do you ne
katrin [286]

Answer:

7.08

Explanation:

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

  • pH = pka + log\frac{[A^-]}{[HA]}

Where \frac{[A^-]}{[HA]} is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:

  • pka = -log (1.8x10⁻⁴) = 3.74

We<u> input the data</u>:

  • 4.59 = 3.74 + log\frac{[A^-]}{[HA]}

And<u> solve for </u>\frac{[A^-]}{[HA]}:

  • 0.85 = log\frac{[A^-]}{[HA]}
  • 10^{(0.85)}=\frac{[A^-]}{[HA]}
  • \frac{[A^-]}{[HA]} = 7.08
3 0
3 years ago
The diagram shows different forms of thermal energy transfer.
Kaylis [27]

Answer:

Conduction

Explanation:

The handle is touching the pot.

And I got it wrong

4 0
3 years ago
What will happen if Tajmahal will sink in sulphuric acid or in hydrochlorine acid?​
adell [148]

Answer:

it will become yellowish in colour as it is made of marble ...

3 0
3 years ago
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