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2 years ago

How would local residents of a place be affected by changes in the hydrosphere? (please help!!)

2 answers:

8 0

The hydrosphere is needed to support human existence. It provides drinking water, water for agricultural purposes, and food and nutrients from fish and plants.
Without hydrosphere the atmosphere will no longer be able to carry up the evaporated water, as well as pour down the water, leaving the biosphere to die.
It will affect all organisms because they won’t be able to survive with water.

— hope it helps!! :)

lesya692 [45]2 years ago

6 0

Answer:

Humans have impacted the hydrosphere drastically and will only continue to due so based on population needs. Global climate change, water pollution, damming of rivers, wetland drainage, reduction in stream flow, and irrigation have all exerted pressure on the hydrosphere's existing freshwater systems.

Hope it helps :)

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Calculate the grams of sulfur dioxide, SO2, produced when a mixture of 35.0 g of carbon disulfide and 30.0 g of oxygen reacts. W

alekssr [168]

Answer:

58.9g of SO2 is produced

8g of oxygen remains unconsumed

Explanation:

The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:

CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)

Molar mass of CS2 = 76.139 g/mol

Molar mass of O2 = 15.99 g/mol

Molar mass of SO2 = 64.066 g/mol

Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles

Number of moles of O2 = 30g/15.999 g/mol =1.88 moles

From the chemical reaction

1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2

Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2

Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced

thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2

Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining

Or 8g of oxygen

58.9g of SO2 is produced

oxygen is the limiting

4 0

3 years ago

If a Ferrari with an initial velocity of 10m/s accelerates at a rate of 5m/s2 for 3 seconds. What will it’s final velocity be?

Reika [66]

Answer:

Explanation:

From first equation of motion, V = U + at

V = 10 + ( 5×3 )

V = 25 m/s

4 0

2 years ago

You have two solid substances that look the same. What measurements would you take and which tests would you preform to determin

ASHA 777 [7]

Answer:

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5 0

2 years ago

What is the difference between the mass number for carbon- 14 and carbon's atomic number of 12.011 amu

Dennis_Churaev [7]

Great question, but I believe you are mixing up atomic number with mass number. Assuming you are, 12.011 amu is the average mass of a carbon atom. For carbon, it can come in three forms: carbon-12, carbon-13, carbon-14. The number following carbon is the mass number of that particular carbon "isotope". The reason the average is so close to 12 is because carbon-12 is by far the most common, so the average should be (and is) very close to 12. Therefore, 12.011 is a weighted average of all carbon molecules, and carbon-14 is a particular carbon molecule that weighs 14 amu.

7 0

3 years ago

PLEASE HELP I HAVE THIS TEST TOMORROW AM GIVING MORE POINTS THAN I SHOULD

zavuch27 [327]

Answer:

41.11 g of Ca(OH)2.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CaO + H2O —> Ca(OH)2

Next, we shall determine the masses of CaO and H2O that reacted and the mass of Ca(OH)2 produced from the balanced equation. This can be obtained as follow:

Molar mass of CaO = 40 + 16 = 56 g/mol

Mass of CaO from the balanced equation = 1 × 56 = 56 g

Molar mass of H2O = (2x1) + 16

= 2 + 16

= 18 g/mol

Mass of H2O from the balanced equation = 1 × 18 = 18 g

Molar mass of Ca(OH)2 = 40 + 2(16 + 1)

= 40 + 2(17)

= 40 + 34

= 74 g/mol

Mass of Ca(OH)2 from the balanced equation = 1 × 74 = 74 g

From the balanced equation above,

56 g of CaO reacted with 18 g of H2O to produce 74 g of Ca(OH)2.

Finally, we obtained the mass of calcium hydroxide, Ca(OH)2 produced from the reaction of 33 g of CaO and 10 g of H2O. This can be obtained as follow:

From the question given above, we were told that when 33 g of CaO and 10 g of H2O reacted, 2 g of CaO were remaining. This implies that H2O is the limiting reactant and CaO is the excess reactant.

Thus, we shall use the limiting reactant to determine the mass of calcium hydroxide, Ca(OH)2 produced because it will give the maximum yield as all of it is consumed in the reaction.

The limiting reactant is H2O and the mass of calcium hydroxide, Ca(OH)2 produced can be obtained as follow:

From the balanced equation above,

18 g of H2O reacted to produce 74 g of Ca(OH)2.

Therefore, 10 g of H2O will react to produce = (10 × 74)/18 = 41.11 g of Ca(OH)2.

Therefore, 41.11 g of Ca(OH)2 were obtained from the reaction.

3 0

3 years ago