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3 years ago

BRAINLIESTTTT!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Chemistry

2 answers:

ludmilkaskok [199]3 years ago

6 0

Answer: B. Algae are an r-selected species because they have a very rapid growth phase.

Reasoning: r-selected species are those that emphasize high growth rates.

never [62]3 years ago

6 0

The correct answer is option B, that is, algae are an r-selected species because they have a very rapid growth phase.

The species that show high growth rates, usually exploit less-crowded ecological niches and generate many offspring, each of which exhibiting a comparatively low possibility of surviving to adulthood are known as r-selected species.

One of the illustrating features of r-selected species is the production of numerous small offspring succeeded by exponential population growth. They mature quickly and require short gestation periods, and exhibits short life spans.

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olasank [31]

Answer:

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Explanation:

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3 years ago

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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

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3 years ago

A student is asked to identify an element that contains 2 valence electrons and is inert (stable). At which location in this per

krek1111 [17]

Answer:

i think the answer is helium (top right in periodic table)

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3 years ago

For an endothermic reaction, how will the value for Keq change when the temperature is increased?

Nataly [62]

I put the answer <em>C: Keq will increase</em>, on PLATO. Hope this works for you!

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3 years ago

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"A certain group one metal combines with Hydrogen, to form a molecule with the formula XH. The equilibrium bond length of this c

Tasya [4]

Based on the bond length, the molecular weight of the unknown metal X is 23.

<h3>What are ionic compounds?</h3>

Ionic compounds are compounds which are formed between oppositely charged ions.

When hydrogen combines with the group 1 metals, they form ionic compounds.

The group one metals that combine with hydrogen to firm ionic bonds include:

Lithium to give Lithium hydride with equilibrium bond length of 0.239Å.
Sodium to give sodium hydride with equilibrium bond length of 1.896Å.
Potassium to give potassium hydride with equilibrium bond length of 2.321Å

The molecular weight of lithium is 7 g

The molecularweight of sodium is 23

The molecular weight of Potassium is 39.

Therefore, based on the equilibrium bond length, the molecular weight of the unknown metal X is 23.

Learn more about bond length at: brainly.com/question/3405662

5 0

3 years ago