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lord [1]
3 years ago
8

Which or the following solutions would have the highest concentration of hydronium ion and why?

Chemistry
1 answer:
chubhunter [2.5K]3 years ago
8 0
Answer is: <span>C)0.050 M HCl.
Chemical reaction 1: HCl(aq) + H</span>₂O(l) → H₃O⁺(aq) + Cl⁻(aq).
Chemical reaction 2: CH₃COOH(aq) + H₂O(l) ⇄ H₃O⁺(aq) + CH₃COO⁻(aq).
Hydrochloric acid (HCl) is strong acid, which means that concencentration of hydronium ions are equal as concentration of hydrochloric acid:
A) c(HCl) = c(H₃O⁺) = 0.001 M.
C) c(HCl) = c(H₃O⁺) = 0.050 M is greater than A.
Acetic acid (CH₃COOH) is weak acid and does not dissociate completely as HCl, so concentration of hydronium ion in this solutions are very low.
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How many moles of Al would be produced from 20 moles of Al2O3?<br> 2Al2O3<br> -&gt;<br> 4A1 + 302
Evgesh-ka [11]
<h3>Answer:</h3>

\displaystyle 40 \ mol \ Al

<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>Stoichiometry</u>

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

<u>Step 1: Define</u>

[RxN - Balanced] 2Al₂O₃ → 4Al + 3O₂

[Given] 20 mol Al₂O₃

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

[RxN] 2 mol Al₂O₃ → 4 mol Al

<u>Step 3: Stoich</u>

  1. [DA] Set up:                                                                                                     \displaystyle 20 \ mol \ Al_2O_3(\frac{4 \ mol \ Al}{2 \ mol \ Al_2O_3})
  2. [DA] Multiply/Divide [Cancel out units]:                                                         \displaystyle 40 \ mol \ Al

<u>Step 4:Check</u>

<em>Follow sig fig rules and round. We are given 1 sig fig.</em>

Since our final answer already has 1 sig fig, there is no need to round.

4 0
3 years ago
Plz someone help, really struggling
frez [133]

Answer:

22.9 Liters CO(g) needed

Explanation:

2CO(g)     +   O₂(g)    =>    2CO₂(g)

? Liters          32.65g

                 = 32.65g/32g/mol

                 =   1.02 moles O₂

Rxn ratio for CO to O₂ = 2 mole CO(g) to 1 mole O₂(g)

∴moles CO(g) needed = 2 x 1.02 moles CO(g) = 2.04 moles CO(g)

Conditions of standard equation* is STP (0°C & 1atm) => 1 mole any gas occupies 22.4 Liters.

∴Volume of CO(g) = 1.02mole x 22.4Liters/mole = 22.9 Liters CO(g) needed

___________________

*Standard Equation => molecular rxn balanced to smallest whole number ratio coefficients is assumed to be at STP conditions (0°C & 1atm).

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