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suter [353]
3 years ago
9

3. A compound was analyzed and found to contain 9.8 g of nitrogen, 0.70 g of

Chemistry
1 answer:
Nesterboy [21]3 years ago
6 0

Answer:

HNO₃

Explanation:

Data given

Nitrogen = 9.8 g

Hydrogen =  0.70 g

Oxygen = 33.6 g

Empirical formula = ?

Solution:

Convert the masses to moles

For Nitrogen

Molar mass of N = 14 g/mol

                              no. of mole = mass in g / molar mass

Put value in above formula

                          no. of mole = 9.8 g/ 14 g/mol

                          no. of mole = 0.7

                           mole of N = 0.7 mol

For Hydrogen

Molar mass of H = 1 g/mol

                     no. of mole = mass in g / molar mass

Put value in above formula

                     no. of mole = 0.70 g/ 1 g/mol

                      no. of mole = 0.7

mole of H = 0.7 mol

For Oxygen

Molar mass of O = 16 g/mol

                       no. of mole = mass in g / molar mass

Put value in above formula

                       no. of mole = 33.6 g / 16 g/mol

                       no. of mole = 2.1

mole of O = 2.1 mol

Now we have values in moles as below

N = 0.7

H = 0.7

O = 2.1

Divide the all values on the smallest values to get whole number ratio

N = 0.7 / 0.7 = 1

H = 0.7 / 0.7 = 1

O = 2.1 / 0.7 = 3

So all have following values

N = 1

H = 1

O = 3

So the empirical formula will be HNO₃ i.e. all three atoms in simplest small ratio.

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Actual (theoretical): 2.70

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How many molecules are in 85g of silver nitrate?
maksim [4K]
<h3>Answer:</h3>

3.0 × 10²³ molecules AgNO₃

<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
  • Writing Compounds
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

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

<u>Step 1: Define</u>

85 g AgNO₃ (silver nitrate)

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

Avogadro's Number

[PT] Molar Mass of Ag - 107.87 g/mol

[PT] Molar Mass of N - 14.01 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of AgNO₃ - 107.87 + 14.01 + 3(16.00) = 169.88 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 85 \ g \ AgNO_3(\frac{1 \ mol \ AgNO_3}{169.88 \ g \ AgNO_3})(\frac{6.022 \cdot 10^{23} \ molecules \ AgNO_3}{1 \ mol \ AgNO_3})
  2. Multiply/Divide:                                                                                                \displaystyle 3.01313 \cdot 10^{23} \ molecules \ AgNO_3

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

3.01313 × 10²³ molecules AgNO₃ ≈ 3.0 × 10²³ molecules AgNO₃

6 0
2 years ago
A standard solution of FeSCN2+ is prepared by combining 9.0 mL of 0.20 M Fe(NO3)3 with 1.0 mL of 0.0020 M KSCN . The standard so
Xelga [282]

Answer : The equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

Explanation :

First we have to calculate the initial moles of Fe^{3+} and SCN^-.

\text{Moles of }Fe^{3+}=\text{Concentration of }Fe^{3+}\times \text{Volume of solution}

\text{Moles of }Fe^{3+}=0.20M\times 9.0mL=1.8mmol

and,

\text{Moles of }SCN^-=\text{Concentration of }SCN^-\times \text{Volume of solution}

\text{Moles of }SCN^-=0.0020M\times 1.0mL=0.0020mmol

The given balanced chemical reaction is,

Fe^{3+}(aq)+SCN^-(aq)\rightleftharpoons FeSCN^{2+}(aq)

Since 1 mole of Fe^{3+} reacts with 1 mole of SCN^- to give 1 mole of FeSCN^{2+}

The limiting reagent is, SCN^-

So, the number of moles of FeSCN^{2+} = 0.0020 mmole

Now we have to calculate the concentration of FeSCN^{2+}.

\text{Concentration of }FeSCN^{2+}=\frac{0.0020mmol}{9.0mL+1.0mL}=0.00020M

Using Beer-Lambert's law :

A=\epsilon \times C\times l

where,

A = absorbance of solution

C = concentration of solution

l = path length

\epsilon = molar absorptivity coefficient

\epsilon and l are same for stock solution and dilute solution. So,

\epsilon l=\frac{A}{C}=\frac{0.480}{0.00020M}=2400M^{-1}

For trial solution:

The equilibrium concentration of SCN^- is,

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{initial} = 0.00050 M

Now calculate the [FeSCN^{2+}].

C=\frac{A}{\epsilon l}=\frac{0.220}{2400M^{-1}}=9.17\times 10^{-5}M

Now calculate the concentration of SCN^-.

[SCN^-]_{eqm}=[SCN^-]_{initial}-[FeSCN^{2+}]

[SCN^-]_{eqm}=(0.00050M)-(9.17\times 10^{-5}M)

[SCN^-]_{eqm}=4.58\times 10^{-8}M

Therefore, the equilibrium concentration of SCN^- in the trial solution is 4.58\times 10^{-8}M

5 0
3 years ago
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