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3 years ago

Question 8

Chemistry

1 answer:

Mice21 [21]3 years ago

5 0

An element I believe

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Molecular distances are given in picometers, where 1 pm is equivalent to 1 x 10-12 meter.If the distance between carbon atoms in

klio [65]

Answer:

Distance = 154 pm

Explanation:

Given that,

The distance between carbon atoms in diamond is $1.54\times 10^{-8}\ cm$

We need to find the distance in pm.

We know that,

$1\ pm=10^{-12}\ m$

$1.54\times 10^{-8}\ cm=1.54\times 10^{-8}\times 10^{-2}\ m\\\\=1.54\times 10^{-10}\ m$

Multiplying and dividing by 10⁻².

$d=1.54\times 10^{-10}\times \dfrac{10^{-2}}{10^{-2}}\\\\=154\times 10^{-12}\ m\\\\=154\ pm$

So, the distance is 154 pm.

8 0

3 years ago

Heat

Kazeer [188]

Answer:

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Explanation:

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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

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