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Ne4ueva [31]
4 years ago
13

In which process are glucose and oxygen the reactants?

Chemistry
2 answers:
Alenkasestr [34]4 years ago
7 0
The answer is B. Cellular Respiration, for sure.
LenaWriter [7]4 years ago
5 0
The answer is B. I took this test a few days ago.
You might be interested in
What is the mass of Al₂O3 that will contain 10 kg of aluminium?​
Oksanka [162]

Answer:

1 mole of Al2O3 = 102 grams

1 mole of Al2 = 54 grams

102 grams of Al2O3 contains = 54 gram of Al2

10kg of Al2O3 contains = (54/102)*10000g Al2

= 5294.11 g Al2 or 5.29411 kg

5 0
3 years ago
Read 2 more answers
Consider the following reaction:
iren [92.7K]

Answer:

A. ΔG° = 132.5 kJ

B. ΔG° = 13.69 kJ

C. ΔG° = -58.59 kJ

Explanation:

Let's consider the following reaction.

CaCO₃(s) → CaO(s) + CO₂(g)

We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.

ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)

where,

n: moles

ΔH°f: standard enthalpy of formation

ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))

ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)

ΔH° = 178.3 kJ

We can calculate the standard entropy of the reaction (ΔS°) using the following expression.

ΔS° = ∑np . S°p - ∑nr . S°r

where,

S: standard entropy

ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))

ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)

ΔS° = 160.6 J/K. = 0.1606 kJ/K.

We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.

ΔG° = ΔH° - T.ΔS°

where,

T: absolute temperature

<h3>A. 285 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ

<h3>B. 1025 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ

<h3>C. 1475 K</h3>

ΔG° = ΔH° - T.ΔS°

ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ

5 0
3 years ago
Can sombody pls help me with this​
luda_lava [24]
Wouldn’t it be half of each? For 36 I guess is 18 and 54 will be 27, (NOT SURE)
6 0
3 years ago
how the various concentrations of acid will affect the amount of limestone that is dissolved from each jar.
schepotkina [342]

The explanation of the how the various concentrations of acid will affect the amount of limestone has been given below.

Effects of acid rain on limestone:-

  • When an acid combines with a carbonate, it produces carbon dioxide as a gas and forms a salt that is soluble in the carbonate and acid's water.
  • There are several gases in the atmosphere that can dissolve in precipitation such as rain and snow.
  • Some may produce acids in rain water, such as carbonic acid, sulfuric acid, and nitric acid.
  • Because the concentration is modest, the rain is not highly acidic, but it is acidic enough to react with the carbonates that make up limestone.

Thus we discussed the affects of acid rain on limestones above.

Learn more about Acid Rain here:

brainly.com/question/718250

#SPJ10

8 0
2 years ago
identify the reagents you would use to convert each of the following compounds into pentanoic acid: (a) 1-pentene (b) 1-bromobut
Morgarella [4.7K]

a)BH3.THF is used to convert 1-pentane to pentanoic acid and b)NaCN is used to convert Bromobutane to pentanoic acid.

a) The conversion of 1-pentane to pentanoic acid using BH3, also known as hydroboration-oxidation, is a two-step reaction involving the reaction of 1-pentane with borane (BH3), followed by oxidation of the resulting 1-pentylborane with hydrogen peroxide or other oxidizing agents.

In the first step, 1-pentane reacts with borane (BH3) to form 1-pentylborane, through a process known as hydroboration. This reaction is catalyzed by a Lewis acid, such as aluminum chloride, and proceeds via a hydride transfer from the borane to the 1-pentane.

In the second step, the 1-pentylborane is oxidized to pentanoic acid using hydrogen peroxide (H₂O₂) or other suitable oxidizing agents. The oxidation is catalyzed by an acid, such as hydrochloric acid (HCl), and proceeds via a proton transfer from the 1-pentylborane to the hydrogen peroxide. The end result is the conversion of 1-pentane to pentanoic acid.

The overall chemical reaction for the conversion of 1-pentane to pentanoic acid using borane (BH₃) and hydrogen peroxide (H₂O₂) is as follows:

1-pentane + BH₃ + H₂O₂ → pentanoic acid + H₂O + BH₂

b)The conversion of 1-Bromo butane to pentanoic acid using sodium cyanide (NaCN) proceeds via a nucleophilic substitution reaction. The reaction mechanism involves the following steps:

1. Attack of the nucleophile, NaCN, on the carbon atom of 1-Bromo butane to form a tetrahedral intermediate.

2. Loss of a proton from the tetrahedral intermediate to form a carbanion.

3. Protonation of the carbanion by water (or another proton source) to form pentanoic acid.

The overall reaction can be represented as follows:

1-Bromo butane + NaCN → Pentanoic Acid + NaBr

To know more about reagents, click below:

brainly.com/question/26283409

#SPJ4

3 0
2 years ago
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