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irinina [24]
2 years ago
6

Why is cellular respiration essential for homeostasis? (4 points)

Chemistry
1 answer:
omeli [17]2 years ago
3 0
It is to create energy that the cell can use. it takes in oxygen to break down food molecules to get cellular energy for cell function
You might be interested in
What mass of silver chloride can be produced from 1.82 l of a 0.176 m solution of silver nitrate?
Fofino [41]
Not sure as don't know ratios, I think it could be 45.93g but don't take my word for it, it could be wrong.


The equations you need are moles = concentration x volume

and mass = moles x formula mass
5 0
3 years ago
An atom of sodium can transfer its one valence electron to an atom of...
seropon [69]

Answer:

I believe it is B,fluorine to complete the octet rule

Explanation:

4 0
3 years ago
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What is a physical intensive property?
Len [333]

An intensive property is a property that does not change depending on how much mass of it you are considered.  An example of an intensive property is density.  No matter how much water you examine, the density of the sample will be 1g/cm³.

6 0
3 years ago
Two iron rods are placed horizontally, touching end-to-end. One end of one rod is heated. After some time, the other rod also fe
Genrish500 [490]

Answer:

Electrons conducted heat

Explanation:

Iron is a metallic compound. One property of metallic compounds are that they have many loose electrons.

Consider the two rods to be a singular rod, since they touch:

An uncountable number of electrons at one end of the rod are heated, so they gain kinetic energy. Those electrons then collide with other electrons, which energize those as well (although less), which is equal to adding heat. This is a pattern that will flow all the way through the rod.

8 0
2 years ago
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A second-order reaction has a rate law: rate = k[a]2, where k = 0.150 m−1s−1. If the initial concentration of a is 0.250 m, what
Cerrena [4.2K]

Rate law for the given 2nd order reaction is:

Rate = k[a]2

Given data:

rate constant k = 0.150 m-1s-1

initial concentration, [a] = 0.250 M

reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s

To determine:

Concentration at time t = 300 s i.e. [a]_{t}

Calculations:

The second order rate equation is:

1/[a]_{t} = kt +1/[a]

substituting for k,t and [a] we get:

1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M

1/[a]t = 49 M-1

[a]t = 1/49 M-1 = 0.0204 M

Hence the concentration of 'a' after t = 5min is 0.020 M



7 0
3 years ago
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