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schepotkina [342]
3 years ago
13

What is the mass in grams of 1.00 x 10 24 atoms of Mn?

Chemistry
2 answers:
hodyreva [135]3 years ago
8 0
A.
Sana po makatulong sagot ko hehe
Alex17521 [72]3 years ago
7 0
<h3>Answer:</h3>

91.2 g Mn

<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
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

[Given] 1.00 × 10²⁴ atoms Mn

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

Avogadro's Numer

[PT] Molar Mass of Mn - 54.94 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                      \displaystyle 1.00 \cdot 10^{24} \ atoms \ Mn(\frac{1 \ mol \ Mn}{6.022 \cdot 10^{23} \ atoms \ Mn})(\frac{54.94 \ g \ Mn}{1 \ mol \ Mn})
  2. [DA] Multiply/Divide [Cancel out units]:                                                           \displaystyle 91.2321 \ g \ Mn

<u>Step 4: Check</u>

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

91.2321 g Mn ≈ 91.2 g Mn

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Answer:

A). The complementary shapes of an enzyme and a substrate.

Explanation:

The Lock-and-key mechanism was proposed by Emil Fischer for the first time and characterized as the metaphor which helps in elucidating the specificity of the enzymatic reactions. In this metaphor, the lock is described as the enzyme while 'key' is characterized as the substrate which the enzyme acts upon. If the key is not appropriately sized, it will not fit into the active site i.e. the keyhole of the lock or enzyme and reaction will not take place. Thus, <u>option A</u> is the correct answer.

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What rusts a nail faster, coke,viniger, or water
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Water, <span>rusts a nail faster</span>
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How many moles of water were lost if the amount of water lost was 0.456 grams? Do not include units and assume three significant
nika2105 [10]
<h3>Answer:</h3>

0.0253 mol H₂O

<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
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 0.456 g H₂O (water)

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

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Convert</u>

  1. [DA] Set up:                                                                                                     \displaystyle 0.456 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})
  2. [DA] Multiply/Divide [Cancel out units]:                                                          \displaystyle 0.025305 \ mol \ H_2O

<u>Step 4: Check</u>

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

0.025305 mol H₂O ≈ 0.0253 mol H₂O

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The reaction of 5.40 g of carbon with excess O2 yields 13.6 g of CO2. What is the percent yield of this reaction?
vazorg [7]
C + O2= CO2
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n =  \frac{13.6}{12 + 16 \times 2} \\ n =  \frac{13.6}{44}  \\ n = 0.31 \: mol \: of \: carbon \: dioxide
CO2 is limit
5.4-3.72= 1.68 g of C is excess
5.4 g = 100%
3.72 g = x
x=68.9 %
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The uranium within these items is radioactive and should be treated with care. Uranium's most stable isotope, uranium-238, has a half-life of about 4,468,000,000 years. It decays into thorium-234 through alpha decay or decays through spontaneous fission.

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