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3 years ago

What is the best solution to groundwater depletion?

1 answer:

3 0

Ground water depletion is one of the major problems. The best solution is to conserve water that we are having currently and to recycle those wasted water. water conservation can be done in many ways such as reduce the level of pumping the ground water, avoiding the use of chemicals, use of natural fertilizers, rain water saving, pumping water to the tanks from stored rain water in the ground, way of handling the water consuming jobs.

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A force has to have which two Factors

rjkz [21]

Hello!
The answer is D Magnitude and Direction.
Hope This Helps! :)

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3 years ago

An avocado contains 234 Calories, with 21 grams of fat. Calculate the percent of Calories from fat.

tatyana61 [14]

36% of calories from fat

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3 years ago

What are the oxidizing agents, and the reducing agents for 2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g)

kondor19780726 [428]

1.) 2Na(aq)+2H2O(l)→2NaOH(aq)+H2(g),

Na is oxidizing agent and H is reducing agent.

2.) C(s)+O2(g)→CO2(g)

C is an oxidizing agent

3.) 2MnO−4(aq)+5SO2(g)+2H2O(l)→2Mn2+(aq)+5SO2−4(aq)+4H+(aq)

Mn is reducing agent and S is oxidizing agent.

4 0

3 years ago

Read 2 more answers

Please help me ASAP I’ll mark Brainly

sammy [17]

Answer:

1. Vacuole

2. chloroplast

3. Nucleus

4. Plasma membrane - cell membrane

5. Vacuole (same as #1 ?) could be vesicle

Explanation:

5 0

3 years ago

A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an

erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

7 0

3 years ago