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Andrews [41]
3 years ago
15

All living things share common characteristics. They are made up of cells, they respond to change, reproduce, have DNA, and use

energy. What is one other common characteristic of living things? have fingers grow and develop live less than 200 years breathe air through lungs
Chemistry
1 answer:
navik [9.2K]3 years ago
4 0
Answer:
they do respiration,they need nutrition,excretion occurs,growth and development happen.
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Shalnov [3]

Answer:

It kinda is

Explanation:

The school system isn't very affective on the young adolcent minds, the crime rate is very high (indicating poverty rates) some drug smmuglers are based in Haiti, and half the city is destroyed from either people or the environment.

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“Fine sediment buries the organism, and pressure eventually squeezes out the liquid and gas parts of the remains, leaving only a
wlad13 [49]
Decomposition I believe. hope this helps

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What is the Ksp of nickel(II) sulfide if the equilibrium concentrations of Ni2+ and S2− in a saturated solution of NiS are each
Ludmilka [50]
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3 years ago
Read 2 more answers
Ethyl iodide (C2H5I) decomposes at a certain temperature in the gas phase as follows: C2H5I(g) → C2H4(g) + HI(g) From the follow
lutik1710 [3]

Answer:

Zero order

Explanation:

Looking at the data we can note a linear dependence between concentration and time.  

Time   Conc.

0          2

15         1.82

30        1.64

48        1.42

75        1.10

In the first 15 min it was consumed 2-1.82=0.18. So the rate is r=\frac{\Delta C}{\Delta t} = \frac{0.18}{15}=0.012

From 15 to 30 min (it has passed 15 min) is consumed 1.82-1.64=0.18, so as in the previous calculation the rate is r=0.012.

From 30 to 48  (it has passed 18 min)the rate is r= \frac{0.22}{18}\approx 0.012  

From 48 to 75  (it has passed 27 min) the rate is r= \frac{0.32}{27}\approx 0.012  

So these results suggest that despite of the ever minor concentration of the reactant the rate is ever the same. Hence the reaction rate could be expressed as r= k^{0} = 0.012 mol L^{-1} min^{-1} that is, the reaction is the zero order respect to C2H5I since it is not depending on concentration of C2H5I.  

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