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Nadusha1986 [10]
2 years ago
5

Identify different hydro-meteorological hazards;

Chemistry
1 answer:
IceJOKER [234]2 years ago
5 0

Hydrometeorological hazards are caused by extreme meteorological and climate events, similar as cataracts, famines, hurricanes, tornadoes, or landslides.

  • Hydrometeorology it’s a branch of meteorology and hydrology that studies the transfer of water and energy between the land face and the lower atmosphere.
  • They regard for a dominant bit of natural hazards and do in all regions of the world, although the frequency and intensity of certain hazards and society’s vulnerability to them differ between regions.
  • Severe storms, strong winds, cataracts, and famines develop at different spatial and temporal scales, but all can come disasters that beget significant structure damage and claim hundreds of thousands of lives annually worldwide. hourly, multiple hazards can do contemporaneously or spark slinging impacts from one extreme rainfall event.
  • In addition to causing injuries, deaths, and material damage, a tropical storm can also affect in flooding and mudslides, which can disrupt water sanctification and sewage disposal systems, beget overflow of poisonous wastes, and increase propagation of mosquito- borne conditions

Therefore these are the hydro-meteorological hazards.

learn more about hydro-meteorological hazards here:

https://brainly.in/question/1347588

#SPJ10

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Gaseous butane, CH3(CH2)2CH, reacts with gaseous oxygen gas, O2, to produce gaseous carbon dioxide, CO2, and gaseous water, H2O.
weeeeeb [17]

Answer:

Percentage yield of carbon dioxide is 49.9%

Explanation:

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

2CH3(CH2)2CH3 + 13O2 —> 8CO2 + 10H2O

OR

2C4H10 + 13O2 —> 8CO2 + 10H2O

Next, we shall determine the masses of butane and oxygen that reacted and the mass of carbon dioxide produced from the balanced equation. This is illustrated below:

Molar mass of butane C4H10 = (12×4) + (10×1)

= 48 + 10

= 58 g/mol

Mass of C4H10 from the balanced equation = 2 × 58 = 116 g

Molar mass of O2 = 16 × 2 = 32 g/mol

Mass of O2 from the balanced equation = 13 × 32 = 416 g

Molar mass of CO2 = 12 + (16×2)

= 12 + 32

= 44 g/mol

Mass of CO2 from the balanced equation = 8 × 44 = 352 g

Summary:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen to produce 352 g of carbon dioxide.

Next, we shall determine the limiting reactant. This can be obtained as follow:

From the balanced equation above,

116 g of butane reacted with 416 g of oxygen.

Therefore, 34.29 g of butane will react with = (34.29 × 416) / 116 = 122.97 g of oxygen.

From the calculation made above, we can see clearly that only 122.97 g out of 165.7 g of oxygen reacted completely with 34.29 g of butane. Therefore, butane is the limiting reactant and oxygen is the excess reactant.

Next, we shall determine the theoretical yield of carbon dioxide.

In this case, we shall use the limiting reactant because it will give the maximum yield of carbon dioxide as all of it is used up in the reaction.

The limiting reactant is butane and the theoretical yield of carbon dioxide can be obtained as follow:

From the balanced equation above,

116 g of butane reacted to produce 352 g of carbon dioxide.

Therefore, 34.29 g of butane will react to produce = (34.29 × 352) / 116 = 104.05 g of carbon dioxide.

Therefore, the theoretical yield of carbon dioxide is 104.05 g

Finally, we shall determine the percentage yield of carbon dioxide as follow:

Actual yield of carbon dioxide = 51.9 g

Theoretical yield of carbon dioxide = 104.05 g

Percentage yield of carbon dioxide =?

Percentage yield = Actual yield /Theoretical yield × 100

Percentage yield of carbon dioxide = 51.9 / 104.05 × 100

Percentage yield of carbon dioxide = 49.9%

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3 years ago
What is the correct name for the ionic compound, MgCl2?
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Answer:

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Is Boyle's law universally true? if not mention one of the its limitations​
murzikaleks [220]
Yes because look in the book dh
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What is the molarity of a stock solution if 60 mL were used to make 150 ml<br>of a .5M solution? ​
olga2289 [7]

The molarity of the stock solution is 1.25 M.

<u>Explanation:</u>

We have to find the molarity of the stock solution using the law of volumetric analysis as,

V1M1 = V2M2

V1 = 150 ml

M1 = 0.5 M

V2 = 60 ml

M2 = ?

The above equation can be rearranged to get M2 as,

M2 = $\frac{V1M1}{V2}

Plugin the values as,

M2 = $\frac{150 \times 0.5}{60}

       = 1.25 M

So the molarity of the stock solution is 1.25 M.

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