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Svetllana [295]
3 years ago
6

A bacterium is in a nasty environment. What sort of reproduction would it use in this situation? Why? How else would it protect

itself? (hint: what were those special capsules called and why were they special?)
Chemistry
1 answer:
lana [24]3 years ago
7 0

The bacteria in nasty environment undergoes multiple fission.

<h3><u>Explanation</u>:</h3>

The bacteria is a unicellular prokaryotic organisms that are found in each and every places of the world. They can survive in extremes of temperatures and pH. They can save themselves through special processes in the extreme climates.

The bacteria undergoes multiple fission in these climates. They cover themselves up with a strong and tough capsule inside which they undergo several Binary fissions. This leads to the formation of multiple cells enclosed with a capsule.

With the return of the favourable climate, the capsule rupture and these newly formed cells come out.

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The most common form if matter in the universe is____?
Rashid [163]

I am almost positive that the answer is: Plasmas. Let me know if im right or not please!! :)

7 0
3 years ago
How many moles of Mg are present in 62.23 grams of Mg? (Show your calculations for full credit.)
mylen [45]
62.23 = 1512.5001499999998 moles
5 0
3 years ago
There are two naturally occurring isotopes of copper. 63cu has a mass of 62.9296 amu. 65cu has a mass of 64.9278 amu. determine
SSSSS [86.1K]
1) You need to use the atomic mass of copper.


You can find it in a periodic table. It is 63.546 amu.


2) The atomic mass is the weigthed mass of the different isotopes.


This is, the atomic mass of one element is the atomic mass of each isotope times its corresponding abundance:


=> atomic mass of the element = abundance isotope 1 * atomic mass isotope 1 + abundance isotope 2 *  atomic mass isotope 2 + ....+abundance isotope n * atomic mass isotope n.


3) The statement tells there are two isotopes so the abundance of one is x and the abundance of the other is 1 - x


=> 63.546 amu = x * 62.9296 amu + (1-x)*64.9278


=> 63.546 = 62.9296x + 64.9278 - 64.9278x


=> 64.9278x - 62.9296 = 64.9278 - 63.546


=> 1.9982x = 1.3818


=> x = 1.3818 / 1.9982 = 0.6915 = 69.15%


=> 1 - x = 1 - 0.6915 = 0.3085 = 30.85%


Answer:


Cu-63 69.15%;


Cu-65 : 30.85%
3 0
3 years ago
What mass (g) of magnesium nitride (Mg3N2) can be made from the reaction of 1.22 g of magnesium with excess nitrogen? __Mg + __N
Citrus2011 [14]
<h3>Answer:</h3>

1.69 g Mg₃N₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table

<u>Stoichiometry</u>

  • Using Dimensional Analysis
  • Reactions RxN
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg + N₂ → Mg₃N₂

[RxN - Balanced] 3Mg + N₂ → Mg₃N₂

[Given] 1.22 g Mg

[Solve] grams Mg₃N₂

<u>Step 2: Identify Conversions</u>

[RxN] 3 mol Mg → Mg₃N₂

[PT] Molar Mass of Mg - 24.31 g/mol

[PT] Molar Mass of N - 14.01 g/mol

Molar Mass of Mg₃N₂ - 3(24.31) + 2(14.01) = 100.95 g/mol

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.22 \ g \ Mg(\frac{1 \ mol \ Mg}{24.31 \ g \ Mg})(\frac{1 \ mol \ Mg_3N_2}{3 \ mol \ Mg})(\frac{100.95 \ g \ Mg_3N_2}{1 \ mol\ Mg_3N_2})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 1.68873 \ g \ Mg_3N_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

1.68873 g Mg₃N₂ ≈ 1.69 g Mg₃N₂

8 0
3 years ago
Choose the formula for calculating percent composition.
kondor19780726 [428]

Answer:

The equation for percent composition is (mass of element/molecular mass) x 100.

please mark me as a brainlist

6 0
3 years ago
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