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3 years ago

Why does algae grow in a lake

Chemistry

2 answers:

pentagon [3]3 years ago

7 0

Answer:

algae will typically grow around the shoreline of a pond or lake because this is where the shallower water is.

Explanation:

skad [1K]3 years ago

5 0

Answer:

Light Exposure and Water Movement. Along with food, algae require the right amount of light to thrive. ... Filamentous algae will typically grow around the shoreline of a pond or lake because this is where the shallower water is.L

Hope that helps :)

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Sterling silver is 92.5% silver and 7.5% copper. Its density is 10.25 g/cm3. A sterling silver pendant is added to a graduated c

Mariulka [41]

From the information given:

The volume of the graduated cylinder = 50.0 mL
when a sterling silver pendant is added, the volume increases to = 61.3 mL

∴

The volume of the sterling silver pendant is:

= 61.3 mL - 50.0 mL

= 11.3 mL

Since, 1 mL = 1cm³

Then;

11.3 mL = 11.3 cm³

the density of the sterling silver = 10.25 g/cm³

Using the relation for Density; i.e.

$\mathbf{Density = \dfrac{mass}{volume}}$

$\mathbf{10.25 \ g/cm^3= \dfrac{mass}{11.3 \ cm^3}}$

mass = 10.25 g/cm³× 11.3 cm³

mass of the sterling silver = 115.825 grams

Recall that sterling silver has:

92.5% silver and;
7.5% copper

∴

The mass of the copper contained in the sterling silver pendant can be calculated as:

$\mathbf{= \dfrac{115.825 \ g \times 7.5}{100}}$

= 8.687 grams

Therefore, we can conclude that the mass of the copper contained in the sterling silver pendant is 8.687 grams

Learn more about the relation between Density, Mass, and Volume here:

brainly.com/question/24386693?referrer=searchResults

4 0

2 years ago

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$