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Troyanec [42]
3 years ago
11

Question 8

Chemistry
1 answer:
Masteriza [31]3 years ago
6 0
AgNO₃ and Cu(NO₃)₂  are dissolved in water
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Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O
Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = \frac{15.0 g}{40 g/mol}=0.375 mol

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

Molarity=\frac{Moles}{Volume(L)}

n=0.250 M\times 0.150 L=0.0375 mol

NaOH+HNO_3\rightarrow NaNO_3+H_2O

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

\frac{1}{1}\times 0.0375 mol of NaOH

Moles of NaOH left unreacted  in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

1\times 0.3375 moles=0.3375 mol of hydroxide ion

The molarity of hydroxide ion in solution ;

=\frac{0.3375 mol}{0.150 L}=2.25 M

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

3 0
3 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<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  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

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

0.424086 J/g °C ≈ 0.424 J/g °C

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