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Vinil7 [7]
3 years ago
7

What is the main product of cellular respiration?

Chemistry
2 answers:
Korvikt [17]3 years ago
6 0

Answer:

The main products are ATP, carbon dioxide, and water.

Gwar [14]3 years ago
4 0
ATP is the primary....


Most of the steps of cellular respiration take place in the mitochondria. Oxygen and glucose are both reactants in the process of cellular respiration. The main product of cellular respiration is ATP; waste products include carbon dioxide and water.
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Which of the following causes a physical change in the shape of an object?
babunello [35]

Answer:

B

Explanation:

Heating a piece of iron until it glows.

3 0
3 years ago
ch question carries 2 mark. Time Remaining : 00 : 46 : 33 Some oxides are given below. (i)Na2O (ii)NO2 (iii) CO2 (iv) MgO a) Whi
cluponka [151]
<h3>Further explanation</h3>

The basic oxide is an oxide-forming a base solution.

These oxides are mainly from group 1 alkaline and group 2-alkaline earth

If this oxide is dissolved in water it will form an alkaline solution

LO + H₂O --> L(OH)₂ ---> alkaline earth

L₂O + H₂O --> LOH --> alkaline

So the basic oxides  : Na₂O and MgO

Na₂O + H₂O --> NaOH (sodium hydroxide, strong base)

MgO + H₂O --> Mg(OH)₂ (magnesium hydroxide, strong base)

The aqueous solution of CO₂ , obtained by dissolving CO₂ in water

CO₂ + H₂O --> H₂CO₃ (carbonic acid)

In general, basic oxide is obtained from metal oxide, while acid oxide is obtained from non-metal oxide

6 0
3 years ago
What is one substance that is 100x more acidic than baking soda
daser333 [38]

Answer:

An acid is a substance that donates hydrogen ions. Because of this, when an acid is dissolved in water, the balance between hydrogen ions and hydroxide ions is shifted. Now there are more hydrogen ions than hydroxide ions in the solution. This kind of solution is acidic.

Explanation:

8 0
3 years ago
The equilibrium constant for the gas phase reaction
stellarik [79]
The equilibrium constant is found by [product]/[reactant]

If the equilibrium constant is very small, such as 4.20 * 10^-31, then that means at equilibrium there is very little product and a lot of reactant.
And likewise, if there is a lot of product formed, and very little reactant, then the K value will be very large, which tells us that it is predominantly product.

At equilibrium, for any reaction, there will always be some reactant and some product present. There cannot be zero reactant or zero product. Also keep in mind that the equilibrium constant is dependent on temperature.

At equilibrium, for your reaction, it is predominantly reactants.
6 0
3 years ago
Air at 7S°F and14.6 though a cfm(cubic ft per minute) and the The flow rate is 48000 psia flows though a rectangular duct of Ix2
IgorLugansk [536]

Explanation:

The given data is as follows.

Air is at 75^{o}F and 14.6 psia.

\varepsilon = 0.00015 ft,     Flow rate, (Q) = 48000 ft^{3}/m

(a)  Formula to calculate hydraulic radius (r_{H}) is as follows.

              r_{H} = \frac{\text{free flow area}}{\text{wet perimeter}}

                          = \frac{2 \times 1}{2(1) + 2(2)}

                          = \frac{1}{3} ft

Formula for equivalent diameter is as follows.

                     D_{eq} = 4 \times r_{H}

                                    = 4 \times \frac{1}{3} ft  

                                    = \frac{4}{3} ft

(b)    Formula for velocity floe is as follows.

                         Q = VA

                     V = \frac{Q}{A}

                        = \frac{48000}{2 \times 1} ft/min

                        = 24000 ft/min

(c)   Formula to calculate Reynold's number is as follows.

         R_{e} = \frac{D \times V \times \rho}{\mu}

                   = \frac{\frac{4}{3} \times 24000 \times 0.0744}{0.0443}  (as \rho = 0.0744 lb/ft^{3} and \mu = 0.0443 lb/ft. hr)

                   = 53742.66 hr/min  

As 1 hr = 10 min. So, 53742.66 hr/min \times \frac{60 min}{1 hr}

                            = 3224559.6

(d)   Formula to calculate pressure drop (\Delta P) is as follows.

              \frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}

Putting the given values into the above formula as follows.

               \frac{\Delta P}{L} = \frac{4f \rho V^{2}}{2Dg_{c}}

                      = \frac{4 \times 0.00015 \times 100 \times 0.0744 \times (24000)^{2}}{2 \times \frac{4}{3} \times {4}{3}}

                      = 6.238 lb/ft^{2}

5 0
3 years ago
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