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Elza [17]
2 years ago
9

Li Ming thinks that chlorophyll is a reactant of photosynthesis, but her classmate Julian disagrees. Who is correct, and why?

Chemistry
2 answers:
Eddi Din [679]2 years ago
7 0

Julian is correct because chlorophyll is neither used up nor formed in the chemical reaction of photosynthesis.

Explanation:

Julian is correct because chlorophyll is neither used up nor formed in the chemical reaction of photosynthesis. The reactants in photosynthesis are carbon dioxide and water.

  • Photosynthesis is the process by which green plants manufacture their food in the presence of sunlight.
  • In photosynthesis carbon dioxide combines with water to produce glucose and oxygen.
  • Chlorophyll in plant is a green pigment that is used to trap sunlight during photosynthesis.

Learn more:

Photosynthesis brainly.com/question/2761166

#learnwithBrainly

Lisa [10]2 years ago
3 0

Answer:

D)

Explanation:

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Explanation:

Ions form when atoms gain or lose electrons. This is so that they form a full outer shell of electrons. When an atom gains electrons it becomes a negative ion, because electrons are negatively charged. For example, all halogens (group 7 or 17) form negative ions as they gain an electron forming a 1- charge. When an atom loses electrons it becomes a positive ion, as it is losing some negative charge from the electrons. This would be for example, alkali metals (group 1) which lose an electron to form a positive ion with a 1+ charge, (ALL metals form positive ions).

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Describe how to test your unknown salt mixture for the presence of<br>Na3PO4.12H2O.​
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Na3PO4*12H2O + BaCl2*2H2O = Ba3(PO4)2 + NaCl + H2O

add barium chloride to your Na3PO4.12H2O a white precipitate of Ba3(PO4)2 will be formed wrt salt(NaCl) and water(H20) if Na3PO4.12H2O. will be there.

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3 years ago
Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---&gt; CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu
Nitella [24]

ΔH = +438 kJ  

We have three equations:  

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ  

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ  

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ  

From these, we must devise the target equation:  

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?  

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.  

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.  

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ  

Equation (V) has 1N₂ on the right, and that is not in the target equation.  

You need an equation with 1N₂ on the left.  

<em>Reverse Equation (III).</em>  

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ  

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.  

You need ³/₂O₂ on the left.  

Multiply <em>Equation (II) by three</em>.  

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ  

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.  

When you add equations, you add their Δ<em>H</em> values.  

_______________________________________

We get the target equation (IV):  

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂;                                    ΔH = +  92 kJ  

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ  

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O;                             ΔH =   -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O;          ΔH =  +438 kJ  


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Answer:

B

Explanation:

Because it is viewed in a different place

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