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Cloud [144]
2 years ago
6

You have 16.0 g of some compound and you perform an experiment to remove all of the oxygen, 11.2 g of iron is left. What is the

empirical formula of this compound?
Chemistry
1 answer:
Makovka662 [10]2 years ago
3 0

The empirical formula of this compound is Fe_2O_3

<h3>Empirical formula </h3>

To calculate the empirical formula of a compound, the value of moles of each element is needed.

As we have the information of the mass value, we will use the molar mass expression, which corresponds to:

MM_O = 16g/mol\\MM_Fe = 55.8g/mol

                                              MM = \frac{m}{mol}

  • O

                                                   16 = \frac{4.8}{x}

                                                   x = 0.3mol

  • Fe

                                                    55.8=\frac{11.2}{x}\\x = 0.2

As the value of the empirical formula must be an integer, simply multiply the two values ​​by a common factor:

                                                O = 0.3 \times 10 = 3\\Fe = 0.2 \times 10 = 2

                                                       Fe_2O_3

So, the empirical formula of this compound is Fe_2O_3.

Learn more about empirical formula: brainly.com/question/1247523

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What is the difference between a pure solvent and a solution? Do they have the same physical properties?
pychu [463]

Answer:

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5 0
3 years ago
Electroplating is a way to coat a complex metal object with a very thin (and hence inexpensive) layer of a precious metal, such
8_murik_8 [283]

Answer:

0.0164 g

Explanation:

Let's consider the reduction of silver (I) to silver that occurs in the cathode during the electroplating.

Ag⁺(aq) + 1 e⁻ → Ag(s)

We can establish the following relations.

  • 1 A = 1 C/s
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The mass of silver deposited when a current of 0.770 A circulates during 19.0 seconds is:

19.0s \times \frac{0.770c}{s} \times \frac{1mole^{-} }{96,468C} \times \frac{1molAg}{1mole^{-}} \times \frac{107.87g}{1molAg} = 0.0164 g

5 0
3 years ago
How many grams of silver nitrate are needed to react with 156.2g of sodium sulfide to produce 595.8g of silver sulfide and 340.0
qaws [65]
Before proceeding, we should write the reaction equation to better understand what is happening:
2AgNO₃ + Na₂S → Ag₂S + 2NaNO₃

Now, we may apply the law of conservation of mass, due to which the total mass before a chemical reaction is equivalent to the total mass after a chemical reaction. Therefore:
Mass of silver nitrate + mass of sodium sulfide = mass of silver sulfide + mass of sodium nitrate

Mass of silver nitrate + 156.2 = 595.8 + 340
Mass of silver nitrate = 779.6 grams
3 0
3 years ago
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Angelina_Jolie [31]

Explanation:

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3. The shells closer to the nucleus are ​smaller​ and can hold ​less​ electrons.

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6. Once the second shell is full, the third shell begins to fill.

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8 0
3 years ago
You wish to prepare an HC2H3O2 buffer with a pH of 4.24. If the pKa of is 4.74, what ratio of C2H3O2 /HC2H3O2 must you use?
LenaWriter [7]

To solve this problem, we can use the Henderson-Hasselbalch Equation which relates the pH to the measure of acidity pKa. The equation is given as:<span>

<span>pH = pKa + log ([base]/[acid])                ---> 1</span></span>

Where,

[base] = concentration of C2H3O2 in molarity or moles

<span>[acid] = concentration of  HC2H3O2 in molarity or moles</span>

 

For the sake of easy calculation, let us assume that:

[base] = 1

[acid] = x

<span>
Therefore using equation 1,
4.24 = 4.74 + log (1 / x) 

<span>log (1 / x) = - 0.5

1 / x = 0.6065 </span></span>

x = 1.65<span>

The required ratio of C2H3O2 /HC2H3O2 <span>is 1:1.65 or 3:5. </span></span>
3 0
3 years ago
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