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Sonbull [250]
3 years ago
5

what is molarity of a sodium hydroxide solution made by combining 2.0 L of 0.60 M NaOH with 495 mL of 3.0 M NaOH? Assume the vol

umes of the two solutions to be additive
Chemistry
1 answer:
hammer [34]3 years ago
8 0

Answer:

Molarity of the sodium hydroxide solution is 1.443 M/L

Explanation:

Given;

0.60 M concentration of NaOH contains 2.0 L

3.0 M concentration of NaOH contains 495 mL

Molarity is given as concentration of the solute per liters of the solvent.

If the volumes of the two solutions are additive, then;

the total volume of NaOH = 2 L + 0.495 L = 2.495 L

the total concentration of NaOH = 0.6 M + 3.0 M = 3.6 M

Molarity of NaOH solution = 3.6 / 2.495

Molarity of NaOH solution = 1.443 M/L

Therefore, molarity of the sodium hydroxide solution is 1.443 M/L

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Consider the hypothetical reaction 4A + 3B → C + 2D Over an interval of 3.00 s the average rate of change of the concentration o
pogonyaev

Answer:

Final concentration of C at the end of the interval of 3s if its initial concentration was 3.0 M, is 3.06 M and if the initial concentration was 3.960 M, the concentration at the end of the interval is 4.02 M

Explanation:

4A + 3B ------> C + 2D

In the 3s interval, the rate of change of the reactant A is given as -0.08 M/s

The amount of A that has reacted at the end of 3 seconds will be

0.08 × 3 = 0.24 M

Assuming the volume of reacting vessel is constant, we can use number of moles and concentration in mol/L interchangeably in the stoichiometric balance.

From the chemical reaction,

4 moles of A gives 1 mole of C

0.24 M of reacted A will form (0.24 × 1)/4 M of C

Amount of C formed at the end of the 3s interval = 0.06 M

If the initial concentration of C was 3 M, the new concentration of C would be (3 + 0.06) = 3.06 M.

If the initial concentration of C was 3.96 M, the new concentration of C would be (3.96 + 0.06) = 4.02 M

3 0
3 years ago
one reaction that produces hydrogen gas can be represented by the unbalanced chemical equation Mg(s)+HCI(aq) -> MgCI(aq)+H2(g
Sonbull [250]
<h3>Answer:</h3>

128 g HCl

<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>

  • Reaction Mole Ratios
  • Using Dimensional Analysis
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Unbalanced] Mg (s) + HCl (aq) → MgCl (aq) + H₂ (g)

↓

[RxN - Balanced] 2Mg (s) + 2HCl (aq) → 2MgCl (aq) + H₂ (g)

[Given] 3.25 mol Mg

[Solve] x g HCl

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

[RxN] 2 mol Mg → 2 mol HCl

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

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of HCl - 1.01 + 35.45 = 36.46 g/mol

<u>Step 3: Stoich</u>

  1. [S - DA] Set up:                                                                                                 \displaystyle 3.25 \ mol \ Mg(\frac{2 \ mol \ HCl}{2 \ mol \ Mg})(\frac{36.46 \ g \ HCl}{1 \ mol \ HCl})
  2. [S - DA] Multiply/Divide [Cancel out units]:                                                    \displaystyle 127.61 \ g \ HCl

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

127.61 g HCl ≈ 128 g HCl

3 0
2 years ago
I need help solving this chemistry question ​
Arturiano [62]

Answer:

I think the answer is 22.2

Explanation: What i DID was adding 12.0 + 10.0 and than that gave me 22 the I had added the to 0.200 and that how i got 22.2. Sorry if i got is wrong. :(

3 0
3 years ago
Magnesium oxide does not readily decompose into magnesium and oxygen. The reaction is shown below. MgO(s) + 601.7 kJ mc017-1.jpg
galben [10]
TEMPERATURE IS THE ANSWER 
5 0
3 years ago
Read 2 more answers
A chemist weighed out 20.7 g of sodium . Calculate the number of moles of sodium she weighed out
Pachacha [2.7K]

Answer:

about 0.9 mol

Explanation:

there are 22.990 g/mol of Na

20.7/22.99 = 0.900391 mol

about 0.9 mol

8 0
3 years ago
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