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3 years ago

How many iron atoms are in Fe2O3?

Chemistry

1 answer:

ira [324]3 years ago

7 0

Answer:

2 iron atoms in Fe2O3

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A student is given a solution of phenol red of unknown concentration. Solutions of phenol red are bright pink under basic condit

azamat

The hydrogen ion concentration, [H ], in the solution of phenol red solution of pH 11.20 is 6.3 X $10^{-12}$ moles/litre.

Explanation:

Data given:

basic solution of phenol red is given as it has pH of 11.2

hydrogen ion concentration = ?

the hydrogen ion concentration is calculated by using the following formula:

pH = - $log_{10}$ [ $H^{+}$ ]

Rearranging the equation:

[ $H^{+}$ ] = $\frac{1}{10^{pH} }$

Putting the values in the above equation:

[ $H^{+}$ ] = $\frac{1}{10^{11.2} }$

[ $H^{+}$ ] = 6.3 X $10^{-12}$ moles/litre

The hydrogen ion concentration is very small and the solution of phenol red is basic in nature. The pH of 11.2 also indicates that solution is basic and teh color of the solution is bright pink.

3 0

4 years ago

What is the atomic mass, in amu, of this atom?

hjlf

Answer:precisely 1/12 the mass of an atom of carbon-12.

Explanation:The carbon-12 (C-12) atom has six protons and six neutrons in its nucleus. In imprecise terms, one AMU is the average of the proton rest mass and the neutron rest mass.

7 0

3 years ago

What is the final temperature in ∘c if the process is isothermal?

VikaD [51]

0.014 mol of gas undergoes the process shown in the figure (Figure 1) .

5 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

PLEASE HELP ASAP

Nady [450]

In the so called rain shadow effect we have interaction between all of the four major Earth spheres. When we have a coastal region where there's a high mountain range, the part of the mountain that is facing the sea will differ a lot from the part of the mountain that is on the other side. The water from the sea evaporates. The water vapor makes the air wet. The warm and wet air masses from the sea will come to the coastline, once they reach the mountain they will start to accumulate as they can not pass through it. As they accumulate rainfall appears. The rainfall contributes to a lush vegetation on this side of the mountain (windward side). The rain shadow effect appears on the leeward side of the mountain, and it mostly gets dry, strong, downward winds. These conditions result in drier climate, much less vegetation, and much increased erosion. Thus we can easily see that we have in this case interaction between the hydrosphere (the sea and the rainfall), the geosphere (the ground, soil, rocks), biosphere (the vegetation), and atmosphere (the winds, the clouds).

7 0

4 years ago