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e-lub [12.9K]
3 years ago
12

Oxygen is important in the oxidation of glucose because it ____________________________. Oxygen is important in the oxidation of

glucose because it ____________________________. can be oxidized to form carbon dioxide gives off energy for various activities donates electrons to make water is highly electronegative
Chemistry
1 answer:
Vlad [161]3 years ago
3 0

can be oxidized to form carbon dioxide

Explanation:

Oxygen is important in the oxidation of glucose because it can be oxidized to form carbon dioxide. Oxidation of glucose involves the reaction of oxygen with glucose in a process called respiration. This gives a product of water, carbon dioxide and energy which is stored as ATP.

  • Oxidation involves the addition of oxygen.
  • Any specie that undergoes oxidation, is a reducing agent and it is said to be oxidized.
  • Oxygen is oxidized to form carbon dioxide and water.

Learn more:

Respiration and photosynthesis brainly.com/question/3437832

#learnwithBrainly

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Determine the possible traits of the calves if:
White raven [17]

Answer:

1. All red calves i.e. RR

2. All roan calves i.e RW

3. 2 red calves (RR) and two roan calves (RW)

Explanation:

According to this question, a gene coding for fur colour in cattle is involved. Red alleles (R) and white alleles (W) are co-dominant to produce a roan cattle (RW). The possible traits of the following crosses are (see attached punnet square):

1) A red bull (RR) is mated to a red (RR) cow: All red calves i.e. RR

2) A red (RR) bullis mated with white (WW) cow: All roan calves i.e RW

3) A roan bull (RW) is mated with red (RR) cow: 2 red calves (RR) and two roan calves (RW).

7 0
3 years ago
How many grams of carbon dioxide are in 35.6 liters of co2
lina2011 [118]
The grams of carbon  dioxide  that are in 35.6 liters  of Co2 is calculates as below
calculate the  number of moles of CO2

At STP  1 mole  = 22.4 L

what  about  35.6 liters

=   1mole x 35.6  liters/ 22.4 liters = 1.589  moles

mass of CO2 =  moles x molar  mass of CO2

= 1.589 mol x 44 g/mol  =  69.92 grams
3 0
3 years ago
A compound is found to contain 47 percent potassium 14.5 percent carbon and 38.5 percent oxygen by the way the molar mass of the
Anettt [7]
Ewafdwagm,kjcnhgfdvbgkl

4 0
3 years ago
Write a balanced equation and determine Eº for each of the following cells: a) Cr Cr3+||Ni2+|Ni b) (CF|C1,||MnO | Mn?)
Alexxx [7]

Answer:

Explanation:

Step 1: Write both half reactions

Cr / Cr3+ : oxidation  Cr(s) → Cr3+ + 3e-  

Ni2+ / Ni : reduction Ni2+ +2e- → Ni

Step 2: Balance reactions and look up the standard potential for the  half-reactions

2(Cr → Cr3+ + 3e-)    E° ox = 0.74 V

3(Ni2+ +2e- → Ni)     E° red = -0.25 V

2Cr + 3Ni2+ +6e- → 2Cr3+ +6e- + 3Ni

E°cell = E° red + E° ox  = -0.25 + 0.74  = 0.49

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C r 3 + ]/ [ N i 2 + ]

With E° = 0.49 V

b)

Step 1: Write both half reactions

MnO4-/ Mn2+ (redution)   MnO4-  +8H+ +5e- ⇔ Mn2+ +4H2O

Cf ⇔ Cf2+ +2e-   (oxidation)   Cf ⇔ Cf2+ +2e-

Step 2: Balance reactions and look up the standard potential for the  half-reactions

MnO4-  +8H+ +10-e- ⇔ Mn2+ +4H2O    E° = 1.51 V

5Cf ⇔ 5 Cf2+ +10e-        E° =2.12 V

2 MnO4- + 16H+ + 5Cf ⇔ 2Mn2+ + 8H2O + 5Cf2+

E°cell = E° red + E° ox  = 1.51 + 2.12  = 3.63 V

E = E ° − 0.0257 V /n * ln Q  = E ° − 0.0257 V /n  *l n [ C f2 + ]/ [ Mno4- ]

With E° =3.63 V

7 0
3 years ago
A mole of a substance has a mass in grams that is equal to the molecular mass. For example, a carbon atom has a mass of 12.01 u.
IRINA_888 [86]

Answer:

M(Fe₂O₃) = 159.70 g/mol

M(CO) = 28.01 g/mol

M(Fe) = 55.85 g/mol

M(CO₂) = 44.01 g/mol

Explanation:

We can calculate the molar mass of a compound by summing the molar masses of the elements that form it.

Fe₂O₃

M(Fe₂O₃) = 2 × M(Fe) + 3 × M(O) = 2 × 55.85 g/mol + 3 × 16.00 g/mol = 159.70 g/mol

CO

M(CO) = 1 × M(C) + 1 × M(O) = 1 × 12.01 g/mol + 1 × 16.00 g/mol = 28.01 g/mol

Fe

M(Fe) = 1 × M(Fe) = 1 × 55.85 g/mol = 55.85 g/mol

CO₂

M(CO₂) = 1 × M(C) + 2 × M(O) = 1 × 12.01 g/mol + 2 × 16.00 g/mol = 44.01 g/mol

3 0
3 years ago
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