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victus00 [196]
3 years ago
15

Plzzz help do your best to answer these questions

Chemistry
1 answer:
vovikov84 [41]3 years ago
5 0

The empirical formula is the simplest formula of a chemical compound.

To find the empirical formula, we take the following steps;

  • Divide the percentage by mass of each element by its relative atomic mass.
  • Divide the quotient of each by the lowest value obtained instep 1 above
  • Write the result of step 2 above as the subscript following each atom.

1) O - 88.10/16,      H - 11.190/1

  O - 5.5,               H - 11.19

  O - 5.5/5.5,        H - 11.19/5.5

  O -  1,                 H - 2

Empirical formula = OH2

2) C - 41.368/12  H - 8.101/1,   N - 32.162/14,   O - 18.369/16

   C - 3,               H - 8,           N - 2,                  O - 1

   C - 3/1,            H - 8/1          N - 2/1                 O - 1/1

    C - 3,             H - 8,           N - 2,                   O - 1

Empirical formula = C3H8N2O

To obtain the molecular formula where n = number of atoms of each element;

Molecular weight = 174.204 g/mol

[ 3(12) + 8(1) + 2(14) + 16]n = 174

n= 174/88

n = 2 (to the nearest whole number)

Hence, we have;

[C3H8N2O]2

The molecular formula is C6H16N4O2

3)  C - 19.999/12,  H - 6.713/1,   N - 46.646/14,   O - 26.641/16

    C - 2,                H - 7,            N -  3,                 O - 2

    C - 2/2,            H - 7/2,         N -   3/2,             O - 2/2

    C - 1,                H - 4,            N -  2,                  O - 1

Empirical formula - CH4N2O

brainly.com/question/1363167

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Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions. If 0.212 g of KHP are dissolve
pochemuha

0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.

The balanced neutralization equation is:

NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)

  • Step 1: Calculate the reacting moles of KHP.

0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.

0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol

  • Step 2: Determine the reacting moles of NaOH.

The molar ratio of NaOH to KHP is 1:1.

1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH

  • Step 3: Calculate the molarity of NaOH.

1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.

[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M

0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.

Learn more about titration here: brainly.com/question/4225093

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What causes a gas to act in a non ideal manner?
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CH4 + O2 → CO2 + H2O
Scilla [17]

Answer:

9.8 × 10²⁴ molecules H₂O

General Formulas and Concepts:

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Moles
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Organic</u>

  • Naming carbons

<u>Stoichiometry</u>

  • Analyzing reaction rxn
  • Using Dimensional Analysis

Explanation:

<u>Step 1: Define</u>

[RxN - Unbalanced] CH₄ + O₂ → CO₂ + H₂O

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 130 g CH₄

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

Avogadro's Number

[RxN] 1 mol CH₄ → 2 mol H₂O

[PT] Molar Mass of C: 12.01 g/mol

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

Molar Mass of CH₄: 12.01 + 4(1.01) = 16.05 g/mol

<u>Step 3: Stoichiometry</u>

  1. [DA] Set up conversion:                                                                                   \displaystyle 130 \ g \ CH_4(\frac{1 \ mol \ CH_4}{16.05 \ g \ CH_4})(\frac{2 \ mol \ H_2O}{1 \ mol \ CH_4})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2O}{1 \ mol \ H_2O})
  2. [DA] Divide/Multiply [Cancel out units]:                                                           \displaystyle 9.75526 \cdot 10^{24} \ molecules \ H_2O

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

9.75526 × 10²⁴ molecules H₂O ≈ 9.8 × 10²⁴ molecules H₂O

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2. Why do you think electroplated jewelleries are in demand?<br> Chemistry plsss
Ivan

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Electroplated jeweleries are in demand because firstly they are as shiny and attractive as real jeweleries, they are also light weighted and cost effective.

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