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leonid [27]
3 years ago
15

When 2.50 g of copper reacts with oxygen the copper oxide product has a mass of 2.81 g. what is the simplest formula of the copp

er oxide?
Chemistry
1 answer:
Rainbow [258]3 years ago
7 0
<span>In a chemical reaction, total mass of reactants should be equal to total mass of products. The total mass of copper oxide which has been produced during the reaction is 2.81 grams, at the beginning there were 2.50 grams of copper which leads us to the mass of Oxygen that has been used in the reaction. 2.50 grams of Copper + X grams of Oxygen = 2.81 grams of Copper Oxide, from this equation the total mass of Oxygen is 0.31 grams. Molar mass of Oxygen is 16 gram/mole so in this oxidation reaction 0.0194 mole of Oxygen is used (0.31/16). Molar mass of Copper is 63.546 gram/mole, in the reaction 0.0393 mole of copper used (2.50/63.546). In order to get the simplest formula, smallest number is used for proportioning. (0.0194/0.0194 = 1) 1 Oxygen (0.0393 / 0.0194 = app. 2) 2 Cu. Simplest formula for this reaction is Cu2O, Copper(I) oxide.</span>
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A chemist adds of a M copper(II) fluoride solution to a reaction flask. Calculate the mass in micrograms of copper(II) fluoride
anygoal [31]

The question is incomplete, here is the complete question.

A chemist adds 345.0 mL of a 0.0013 mM (MIllimolar) copper(II) fluoride CuF_2 solution to a reaction flask.

Calculate the mass in micrograms of copper(II) fluoride the chemist has added to the flask. Be sure your answer has the correct number of significant digits.

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Explanation :  Given,

Millimolarity of copper (II) fluoride = 0.0013 mM

This means that 0.0013 millimoles of copper (II) fluoride is present in 1 L of solution

Converting millimoles into moles, we use the conversion factor:

1 moles = 1000 millimoles

So, 0.0013mmol\times \frac{1mol}{1000mmol}=1.3\times 10^{-6}mol

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

We are given:

Moles of copper (II) fluoride solution = 1.3\times 10^{-6}mol

Molar mass of copper (II) fluoride = 101.5 g/mol

Putting values in above equation, we get:

1.3\times 10^{-6}mol=\frac{\text{Mass of copper (II) fluoride}}{101.5g/mol}\\\\\text{Mass of copper (II) fluoride}=(1.3\times 10^{-6}mol\times 101.5g/mol)=1.32\times 10^{-4}g

Converting this into milligrams, we use the conversion factor:

1 g = 1000 mg

So,

\Rightarrow 1.32\times 10^{-4}g\times (\frac{1000mg}{1g})=0.13mg

Therefore, the mass of copper(II) fluoride is, 0.13 mg

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