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Paul [167]
3 years ago
14

15. How many grams are equal to 0.11 mole of copper(1) chromate, Cu2(CrO4)? with work please​

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
7 0

Answer:

26.8g

Explanation:

The formula of the compound given is:

          Cu₂CrO₄

Given:

Number of moles  = 0.11

To find the mass, we use the expression below:

        Mass  = number of moles x molar mass

Molar mass of  Cu₂CrO₄ = 2(63.6) + 52 + 4(16) = 243.2g/mol

Now insert the parameters and solve;

 Mass  = 0.11 x 243.2 = 26.8g

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In a ketone, the carbonyl group is bonded to how many hydrogen atoms? none one two three four
allochka39001 [22]

Answer:

None of carbonyl group is attached to hydrogen atoms.

Explanation:

Ketone is a group in which one carbonyl group is attached to two R ( the alkyl group) .

And the group in which two hydrogen is attached to R is called as aldehyde (formaldehyde).

And when one hydrogen and one alkyl group is attached to R. It is called as acetaldehyde from aldehyde family.

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5 0
2 years ago
How many grams of fluorine are contained in 8 molecules of boron trifluoride?
Lelu [443]
<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

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