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Alexus [3.1K]
3 years ago
7

Sort the functions of life based on whether they are performed by an amoeba or a plant.makes food fromcarbon dioxide,water, and

sunlighthas only offspringwith identicalgenes and traitssurrounds and engulfsfood particlescan have offspringwith differentgenes and traitscan move freely intheir environment
Chemistry
2 answers:
san4es73 [151]3 years ago
8 0

functions of life based on performed by an  a plant:

1. Makes food from carbon dioxide,water, and sunlight.

2. Can have offspring with different genes and traits.

functions of life based on performed by an amoeba:

1. Has only off spring with identical genes and traits surrounds and engulfs food particles.

2. can move freely in their environment.



jeka943 years ago
3 0

Explanation:

Plants are autotrophic organisms which prepare their food with the help of photosynthesis. Plants can have both asexual or multi-sexual reproduction.

Amoeba is a single celled eukaryotic organism. It moves through pseudopodia and it does not have any definite shape. Amoeba can have only asexual reproduction.

The given functions for amoeba and plants are sorted as follows.

Functions of Amoeba:

  • has only offspring with identical genes and traits.
  • surrounds and engulfs food particles.
  • move freely in their environment.

Functions of Plants:

  • makes food from carbon dioxide,water, and sunlight.
  • can have offspring with different genes and traits.

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Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


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