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babunello [35]
3 years ago
5

Will NaBr, CO2, Ca3(PO4), or KZS raise a liquid's boiling point the most when

Chemistry
1 answer:
Vladimir79 [104]3 years ago
6 0

Answer:

K_2S

Explanation:

According to the boiling point elevation law described by the equation \Delta T_b = iK_bb, the increase in boiling point is directly proportional to the van 't Hoff factor.

The van 't Hoff factor for nonelectrolytes is 1, while for ionic substances, it is equal to the number of moles of ions produced when 1 mole of salt dissolves.

NaBr would produce 2 moles of ions per 1 mole of dissolved substance, sodium and bromide ions.

Ca_3(PO_4)_2 is insoluble in water, so it would barely dissociate and wouldn't practically change the boiling point.

K_2S would dissociate into 3 moles of ions per 1 mole of substance, two potassium cations and one sulfide anion.

CO_2 is a gas, it would form some amount of carbonic acid when dissolved, however, carbonic acid is molecular and would yield i value of i = 1.

Therefore, potassium sulfide would raise a liquid's boiling point the most if all concentrations are equal.

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3 years ago
How many molecules are contained in 103.4g of sulfuric acid?
Ierofanga [76]
<h3>Answer:</h3>

1.827 × 10²⁴ molecules H₂S

<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>Chemistry</u>

<u>Compounds</u>

  • Writing Compounds
  • Acids/Bases

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

103.4 g H₂S (Sulfuric Acid)

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

Avogadro's Number

Molar Mass of H - 1.01 g/mol

Molar Mass of S - 32.07 g/mol

Molar Mass of H₂S - 2(1.01) + 32.07 = 34.09 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 103.4 \ g \ H_2S(\frac{1 \ mol \ H_2S}{34.09 \ g \ H_2S})(\frac{6.022 \cdot 10^{23} \ molecules \ H_2S}{1 \ mol \ H_2S})
  2. Multiply:                                                                                                            \displaystyle 1.82656 \cdot 10^{24} \ molecules \ H_2S

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 4 sig figs.</em>

1.82656 × 10²⁴ molecules H₂S ≈ 1.827 × 10²⁴ molecules H₂S

4 0
3 years ago
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