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MakcuM [25]
3 years ago
8

What is the density of a sample with a mass of 5250g and a volume of 4.5mL

Chemistry
1 answer:
vlabodo [156]3 years ago
5 0

Answer:

1166.67

Explanation:

density=mass/volume

density= 5250/4.5

density=1166.67g/mL

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Sophie [7]
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8 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
2 years ago
4. Consider the following half-reactions: MnO4–(aq) + 8H+(aq) + 5e– → Mn+2(aq) + 4H2O(l) NO3–(aq) + 4H+(aq) + 3e– → NO(g) + 2H2O
jeyben [28]

Explanation:

The chemical reaction given in the question is as follows -

MnO₄⁻ (aq) + 8H⁺ (aq) + 5e⁻ → Mn²⁺ (aq) + 4H₂O (l)

NO₃⁻ (aq) + 4H⁺ (aq) + 3e⁻ → NO (g) + 2H₂O (l)

As we know , the value for reduction potential are -

Mn²⁺ = + 1.51  V

NO₃⁻  =  +0.96 V

From , the data given above , the value of the reduction potential of NO₃⁻ is less than the reduction potential of Mn²⁺ .

Hence ,

NO₃⁻  can not oxidize Mn²⁺ .

5 0
3 years ago
what does a set of four quantum numbers tell you about an electron. compare and contrast the locations and properties of two ele
daser333 [38]
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7 0
3 years ago
A radioactive sample has a half-life of 5 hours. What fraction of the sample will be left after 15 hours?​
DedPeter [7]

Explanation:

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6 0
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