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Colt1911 [192]
3 years ago
7

(04.03 HC)

Chemistry
1 answer:
julia-pushkina [17]3 years ago
7 0

The process our cells use to take in food and oxygen is called cellular respiration. Cellular respiration is a set of chemical reactions cells use to change the food we eat, the water we drink, and oxygen from the air we breathe into forms the cell can use as energy.Cellular respiration is the process by which the chemical energy of "food" molecules is released and partially captured in the form of ATP. Carbohydrates, fats, and proteins can all be used as fuels in cellular respiration, but glucose is most commonly used as an example to examine the reactions and pathways involved.

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PLEASE HELP IM DYING OF CHEMISTRY, I NEED HELP FINDING ATOMIC MASS WITH ISOTOPES
Nikolay [14]

Answer:

option A is correct

Explanation:

5 0
3 years ago
In the stratosphere, oxygen (o2) is converted to ozone (o3) by high energy ultraviolet light. when the equation is written and b
Sergio039 [100]
3O₂ = 2O₃ (ozone), so coefficient is 2.
8 0
3 years ago
A galvanic (voltaic) cell consists of an electrode composed of chromium in a 1.0 M chromium(III) ion solution and another electr
svetlana [45]

Answer: The potential of the following electrochemical cell is 1.08 V.

Explanation:

E^0_(Cr^{3+}/Cr)=-0.74V[/tex]

E^0_(Cu^{2+}/Cu)=0.34V[/tex]

The element  with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.

2Cr+3Cu^{2+}\rightarrow 2Cr^{3+}+3Cu

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials, when concentration is 1M.

E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}

E^0=0.34-(-0.74)=1.08V

Thus the potential of the following electrochemical cell is 1.08 V.

6 0
3 years ago
What 2 properties make capillary action possible
Natali5045456 [20]
Cohesion and adhesion
7 0
3 years ago
How many moles of chlorine are used up in a reaction that produces 0.35kg of BCl3
maw [93]

4.48 mol Cl2. A reaction that produces 0.35 kg of BCl3 will use 4.48 mol of Cl2.

(a) The <em>balanced chemical equation </em>is

2B + 3Cl2 → 2BCl3

(b) Convert kilograms of BCl3 to moles of BCl3

MM: B = 10.81; Cl = 35.45; BCl3 = 117.16

Moles of BCl3 = 350 g BCl3 x (1 mol BCl3/117.16 g BCl3) = 2.987 mol BCl3

(c) Use the <em>molar ratio</em> of Cl2:BCl3 to calculate the moles of Cl2.

Moles of Cl2 = 2.987 mol BCl3 x (3 mol Cl2/2 mol BCl3) = 4.48 mol Cl2

4 0
3 years ago
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