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2 years ago

A 20.0 mL sample of hydrochloric acid was neutralized with 58.4 mL of 0.250 M NaOH, in a titration.

Chemistry

1 answer:

erik [133]2 years ago

5 0

Answer: The Answer is 0.73.

Explanation: Solved in the attached picture.

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Consider this mechanism:

zaharov [31]

The answer is intermediate, so E.

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2 years ago

how do you know the state of a compound when writing out the equation. WILL MARK BRAINLIEST PLEASE HELP!!

tester [92]

Answer:

The state of matter of each compound or molecule is indicated in subscript next to the compound by an abbreviation in parentheses. For example, a compound in the gas state would be indicated by (g), solid (s), liquid (l), and aqueous (aq).

Explanation:

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2 years ago

The public is not yet able to purchase cars powered by hydrogen fuel cells because engineers have to determine how the cars perf

Leya [2.2K]

Answer:

It’s B just did it

Explanation:

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3 years ago

the student records the concentration of the stock solution of H2O2 to be 0.950 M. The student proceeds to prepare the reaction

lana [24]

Answer:

0.0400M of KI

Explanation:

Molarity is an unit of concentration defined as the ratio between moles of solute and liters of solution.

When you add 10.0 mL of 0.10M KI and 15.0mL, total volume is:

25.0mL = <em>0.025L of solution</em>

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And moles of KI are:

0.0100L × 0.10M = <em>0.00100 moles of KI</em>

<em />

Thus, molarity is:

0.00100 moles / 0.025L = <em>0.0400M of KI</em>

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3 years ago

1 Na2CO3(aq) + 1 CaCl2(aq) → 1 CaCO3(s) + 2 NaCl(aq) 4. Use the balanced chemical equation from the last question to solve this

LenKa [72]

<h3>Answer:</h3>

0.6 g NaCl

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<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>

[RxN - Balanced] Na₂CO₃ (aq) + CaCl₂ (aq) → CaCO₃ (s) + 2NaCl (aq)

[Given] 0.5 g Na₂CO₃ reacted with excess CaCl₂

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

[RxN] Na₂CO₃ → 2NaCl

Molar Mass of Na - 22.99 g/mol

Molar Mass of C - 12.01 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Cl - 35.45 g/mol

Molar Mass of Na₂CO₃ - 2(22.99) + 12.01 + 3(16.00) = 105.99 g/mol

Molar Mass of NaCl - 22.99 + 35.45 = 58.44 g/mol

<u>Step 3: Stoichiometry</u>

Set up: $\displaystyle 0.5 \ g \ Na_2CO_3(\frac{1 \ mol \ Na_2CO_3}{105.99 \ g \ Na_2CO_3})(\frac{2 \ mol \ NaCl}{1 \ mol \ Na_2CO_3})(\frac{58.44 \ g \ NaCl}{1 \ mol \ NaCl})$
Multiply/Divide: $\displaystyle 0.551373 \ g \ NaCl$

<u>Step 4: Check</u>

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

0.551373 g NaCl ≈ 0.6 g NaCl

5 0

2 years ago