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3 years ago

Select all the correct answers.

Chemistry

2 answers:

balu736 [363]3 years ago

7 0

Answer:photosynthesis by plants

absorption of CO2 from the atmosphere by ocean water

Explanation:

Photosynthesis involves the combination of CO2 with water to form glucose and subsequently complex carbohydrates. Plants constitute a very important sink for carbon IV oxide in nature. Secondly, oceans absorb carbon IV oxide to maintain the concentration of bicarbonate in the water. These are two important natural methods that remove CO2 from the environment.

Archy [21]3 years ago

4 0

Answer:

The natural method of removal of CO2 is

Photosynthesis by plants

Absorption of CO2 from the atmosphere by ocean water

Melting of glacial ice

Explanation:

Respiration of living organism : In this process, Oxygen is consumed(removed) and Carbon dioxide(CO2 ) is formed.So it forms CO2 .

Photosynthesis by plants: Green plant containing. prepare their food in presence of sunlight by using CO2 , water ,Hence CO2 is removed from atmosphere.

Evaporation of ocean water : It increases the level of CO2 in atmosphere. The heating of oceans surface Degassed CO2 to atmosphere.

Absorption of CO2 from the atmosphere by ocean water : the plants under the oceans needs CO2 to carry out photosynthesis .Hence oceans absorbs CO2 from atmosphere .Here, CO2 is removed from atmosphere.

Melting of glacial ice : Glacial ice absorb carbon dioxide because it does not have organism respiration (organisms releasing CO2 ).They capture large amount of CO2 and called CO2 sinks.(Remove CO2)

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Answer:

$\large \boxed{\text{0.0503 g}}$

Explanation:

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Assemble all the data in one place, with molar masses above the formulas and masses below them.

M_r: 39.10 80.41 2.016

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m/g: 5.5 4.04

a) Limiting reactant

(i) Calculate the moles of each reactant

$\text{Moles of K} = \text{5.5 g} \times \dfrac{\text{1 mol}}{\text{31.10 g}} = \text{0.141 mol K}\\\\\text{Moles of HBr} = \text{4.04 g} \times \dfrac{\text{1 mol}}{\text{80.91 g}} = \text{0.049 93 mol HBr}$

(ii) Calculate the moles of H₂ we can obtain from each reactant.

From K:

The molar ratio of H₂:K is 1:2.

$\text{Moles of H}_{2} = \text{0.141 mol K} \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol K}} = \text{0.0703 mol H}_{2}$

From HBr:

The molar ratio of H₂:HBr is 3:2.

$\text{Moles of H}_{2} = \text{0.049.93 mol HBr } \times \dfrac{\text{1 mol H}_{2}}{\text{1 mol HBr}} = \text{0.024 97 mol H}_{2}$

(iii) Identify the limiting reactant

HBr is the limiting reactant because it gives the smaller amount of NH₃.

b) Excess reactant

The excess reactant is K.

c) Mass of H₂

$\text{Mass of H}_{2} = \text{0.024 97 mol H}_{2} \times \dfrac{\text{2.016 g H}_{2}}{\text{1 mol H}_{2}} = \textbf{0.0503 g H}_{2}\\ \text{The mass of hydrogen is $\large \boxed{\textbf{0.0503 g}}$ }$

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