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3 years ago

Which of the following best describes a double-replacement reaction?

Chemistry

1 answer:

Sholpan [36]3 years ago

5 0

What are the answer choices

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Two scientists said to report unusual findings when they shared their results from a recent science experiment. Why is this not

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3 years ago

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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.

Answer : The mass of copper(II) fluoride is, 0.13 mg

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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3 years ago

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The awnser for this should be h

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