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3 years ago

What are the necessary ingredients for photosynthesis

Chemistry

2 answers:

Eduardwww [97]3 years ago

8 0

Answer:

Answer is water, carbon dioxide, and energy, which yields glucose and oxygen, as in the well-known formula: 6 H2O + 6 CO2 + Energy = C6H12O6 + 6 O2. Hope it helps!

Stels [109]3 years ago

3 0

Answer:

Water (H₂O), sunlight, and carbon dioxide (CO₂)

Explanation:

With water, sunlight, and carbon dioxide the plant can perform photosynthesis and arrange the atoms making glucose (C₆H₁₂O₆)

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Answer:A moon

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Students were investigating properties of matter, but wanted to make sure that when they tested these properties they had proper

Mariana [72]

Answer:

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Explanation:

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4 years ago

A general chemistry student found a chunk of metal in the basement of a friend's house. To figure out what it was, she tried the

mr_godi [17]

Answer:

do you have options for this

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3 years ago

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Select the correct answer. One night, an ice storm strikes your town. When you go outside in the morning, your entire front walk

Sever21 [200]

Sucrose will be the most effective. Hence, option D is correct.

<h3>What is sucrose?</h3>

Sucrose is simply the chemical name for sugar, the simple carbohydrate we know and love that is produced naturally in all plants, including fruits, vegetables and even nuts.

When salt or sugar (a solute) is combined with water or ice (a solvent) and is evenly distributed, the freezing point is lowered.

$(C_{12}H_{22}O_{11}$ → $C_{12}H_{22}O_{11})$

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2 years ago

When 40.5 g of Al and 212.7 g of Cl2 combine in the reaction: 2 Al (s) + 3 Cl2 (g) --> 2 AlCl3 (s) c) How many moles of the e

sineoko [7]

Answer:

The answer to your question is 1.49 mol of Cl₂

Explanation:

Data

mass of Al = 40.5 g

mass of Cl₂ = 212.7 g

moles of excess reactant = ?

- Balanced chemical reaction

2 Al(s) + 3Cl₂(g) ⇒ 2AlCl₃

Process

a) Calculate the molar mass of the reactants

molar mass of Al = 2 x 26.98 = 53.96 g

molar mass of Cl₂ = 6 x 35.45 = 212.7 g

b) Calculate the theoretical proportion Al/Cl₂ = 53.96/212.7 = 0.254

Calculate the experimental proportion Al/Cl₂ = 40.5/212.7 = 0.19

As the experimental proportion is lower than the theoretical proportion we conclude that the limiting reactant is Aluminum.

c) Calculate the grams of excess reactant

53.96 g of Al ------------------ 212.7 g of Cl₂

40.5 g of Al ------------------- x

x = (40.5 x 212.7) / 53.96

x = 8614.35 / 53.96

x = 159.64 g of Cl₂

Excess Cl₂ = 212.7 - 159.64

= 53.057 g

d) Calculate the moles of Cl

35.45 g of Cl ----------------- 1 mol

53.057 g of Cl --------------- x

x = (53.057 x 1)/35.45

x = 1.49 mol of Cl₂

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