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Masteriza [31]
3 years ago
12

Populations in the deep-sea vent ecosystem can get energy storage molecules without sunlight. Do you agree or disagree with this

claim? Explain your answer.
Chemistry
1 answer:
Mamont248 [21]3 years ago
5 0

Answer:

Agree

Explanation:

<em>The deep-sea vent ecosystem gets energy storage molecules without sunlight.</em> Life in the vent ecosystem relies on alternate energy through the use of hydrogen sulfide present in the water emanating from the vent.

The producers that form the basis of the food chain are bacteria that use the hydrogen sulfide to create energy-rich molecule in a process known as chemosynthesis.

The equation of the chemosynthesis is as below:

CO_2 + 4H_2S + O_2 --> CH_20 + 4S + 3H_2O

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An intrusion is any body of intrusive igneous rock, formed from magma that cools and solidifies within the crust of the planet. In contrast, an extrusion consists of extrusive rock, formed above the surface of the crust.
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What is true about the ocean temperature
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Answer:

The average temperature of the sea surface is about 20° C (68° F), but it ranges from more than 30° C (86° F) in warm tropical regions to less than 0°C at high latitudes. In most of the ocean, the water becomes colder with increasing depth

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3 years ago
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­­2K + 2HBr → 2 KBr + H2
Inessa [10]

Answer:

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

Explanation:

The limiting reactant is the reactant that gives the smaller amount of product.

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  

            2K  +  2HBr ⟶ 2KBr + H₂

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

3 0
3 years ago
A sample of solid NH4NO3 was placed in an evacuated container and then heated so that it decomposed explosively according to the
dybincka [34]

Answer:

Kp is 2.98  (option c.)

Explanation:

The decomposition is:

NH₄NO₃(s) →  N₂O(g) + 2H₂O(g)

Total pressure at equilibrium is 2.72 bar so, in order to determine Kp we need the partial pressure and we only have total pressure.

According to stoichiometry, nitrogen oxide increase by 1, the partial pressure and water vapor, by 2.

Total pressure is: Partial pressure N₂O + Partial pressure H₂O

2.72 bar = X + 2X → X = 2.72 bar / 3 = 0.91 bar

Partial pressure N₂O = 0.91 bar

Partial pressure H₂O = 1.81 bar

We make the expression for Kp = Partial pressure N₂O . (P. pressure H₂O)²

Kp = 0.91 . 1.81² = 2.98

We do not consider the ammonium nitrate, because it is solid

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Please help answer!!!​
Mazyrski [523]

Answer:

C stays the same ez

Explanation:

7 0
3 years ago
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