Which of the following renewable energy sources can be used for electricity, vehicle fuel, and heat?

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Answer 1

Answer: The answer would be Biomass

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Using the graph below, at what depth does the thermocline begin?A line graph relating ocean depth to water temperature. At 0 meters, temperature is approximately 27°C. At 50 meters, temperature is approximately 24°C. At 100 meters, temperature is approximately 11°C. At 150 meters, temperature is approximately 8°C.

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0 meters
50 meters
100 meters
150 meters

Answers

Answer: The thermocline begins at 100 meters of depth.

Step-to-step explanation:-

Thermocline is a transition oceanic water layer between deep and surface water in which water temperature decreases rapidly with increasing depth.

From the given graph it cam be seen that at 100 meters the thermoline begins such that the temperature drops from $27^(\circ)$ to $24^(\circ)$ .

Hence, the thermocline begins at 100 meters of depth.

Answer:

D. 150 m

Explanation:

Is yogurt alkaline, neutral or acidic?

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Is yogurt alkaline, neutral or acidic?neutral boardering alkaline

in a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution? Use mc018-1.jpg.

Answers

Ans: Volume of stock H2SO4 required = 6.94 ml

Given:

Concentration of stock H2SO4 solution M1 = 18.0 M

Concentration of the final H2SO4 solution needed M2 = 2.50 M

Final volume of H2SO4 needed, V2 = 50.0 ml

To determine:

Volume of stock needed, V1

Explanation:

Use the dilution relation:

$M1V1 = M2V2\n\nV1 = (M2V2)/(M1) \n\nV1 = (2.50 M * 50.0 ml)/(18.0 M) = 6.94 ml$

Hello!

In a school’s laboratory, students require 50.0 mL of 2.50 M H2SO4 for an experiment, but the only available stock solution of the acid has a concentration of 18.0 M. What volume of the stock solution would they use to make the required solution?

We have the following data:

M1 (initial molarity) = 2.50 M (or mol/L)

V1 (initial volume) = 50.0 mL → 0.05 L

M2 (final molarity) = 18.0 M (or mol/L)

V2 (final volume) = ? (in mL)

Let's use the formula of dilution and molarity, so we have:

$M_(1) * V_(1) = M_(2) * V_(2)$