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saw5 [17]
2 years ago
8

Why would it be better to be an r-selected species if the water resources in an area were to become more limited over a short pe

riod of time?
Also, sorry if this is the wrong subject, it was hard to find a fitting one for Environmental Science.
Chemistry
2 answers:
Ulleksa [173]2 years ago
7 0
Hello there!

It would be better to be an r-selected species if the water resources in an area were to become more limited because they tend to reproduce faster than other animals

Hope this helps! :)
Gnoma [55]2 years ago
6 0
An r-selected species reproduces much faster than K-selected species.

r-selected species focuses on maturing and reproducing quickly. r-selected species will probably reproduce when the water supply is there for the short period of time; thus, increasing the chance of the r-selected species of surviving.

K-selected species, on the other hand, focus on raising their young and reproduce later. Since the K-selected species take long to mature before reproducing, water supply may run out before they have a chance of fully maturing; thus, K-selected species have a lower chance of survival.

Hope this helps.
If you need anything more, feel free to comment! Have an awesome day! :)

~Collinjun0827, Junior Moderator
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Element X has the following valence electron configuration: [core]ns2np5. Element M has the following valence electron configura
alex41 [277]

Answer:

The kind of ionic compound formed is MX2.

Explanation:

Element X electron  configuration is represented as [core] ns2np5. The group in the periodic table this element belong to is group 7A.  The element group is called the halogen family. Element X cannot be stated specifically, because the number is represented with n.  Element X will behave as an anions when it react with a metal(cations). Element X has a charge of -1. The element X will gain electron when it bond with a metal. Element X is a non metal . Elements  in this group are fluorine, chlorine, bromine, iodine , astatine, and tennessine . The element X have 7 valency electrons.

Element M electronic configuration is represented as  [core]ns2. The group in the periodic table this element belong to is group 2A . The element  group is called the alkaline earth metals family . Element M will behave as a cation when it bond with a non metal. Element M is a metal , therefore it will likely lose electron to form cations during bonding . The charge of element M is 2+. Element M is positively charged. Elements that belong to this group includes beryllium, magnesium, calcium, strontium, barium and radium.  Element M has 2 valency electrons.

The reaction between this 2 ions will likely form an ionic compound . The element M is the cations while the element X is the anions. The element M will lose 2 electron while 2 atoms of element  X will gain 2 electrons.Element M will lose 2 electron to attain a stable configuration while 2 atoms of element X will gain a single electron each to attain a stable configuration.

M²+ and F- . This will form MX2 when you cross multiply the charge. The kind of ionic compound formed is MX2.

6 0
3 years ago
Write the mechanism of the reaction of methyl benzoate to form methyl 2-nitrobenzoate.
Galina-37 [17]
The question is partially incorrect, because nitration of <span> methyl benzoate results in generation of methyl 3-nitrobenzoate, and not methyl 2-benzoate.

This a because of the present of ester group, which deactivated benzene ring at ortho and para position. Due to this, the electrophile (NO2+) attackes on meta position.

The detailed mechanism is attached below.</span>

8 0
3 years ago
Read 2 more answers
How much heat is required to raise the temperature of 670g of water from 25.7"C to 66,0°C? The specific heat
inna [77]

Answer:

Explanation:

q= mc theta

where,

Q = heat gained

m = mass of the substance = 670g

c = heat capacity of water= 4.1 J/g°C    

theta =Change in temperature=( 66-25.7)

Now put all the given values in the above formula, we get the amount of heat needed.

q= mctheta

q=670*4.1*(66-25.7)

  =670*4.1*40.3

=110704.1

8 0
3 years ago
the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107
oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u)  and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0
3 years ago
6 × 10^9 mm = m. how to convert millimeters to m​
OLga [1]

Answer:

6x10^9x10^-3m=6x10^6m

Explanation:

3 0
3 years ago
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