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yaroslaw [1]
3 years ago
13

How many grams of CaCo3 are required to prepare 50.0g of CaO

Chemistry
1 answer:
olchik [2.2K]3 years ago
7 0
Answer is: 89 g of CaCO₃ is <span>equired to prepare 50.0g of CaO.
</span>Chemical reaction: CaCO₃ → CaO + CO₂.
m(CaO) = 50,0 g.
m(CaCO₃) = ?
n(CaO) = m(CaO) ÷ M(CaO).
n(CaO) = 50,0 g ÷ 56 g/mol.
n(CaO) = 0,89 mol.
from reaction: n(CaCO₃) : n(CaO) = 1 : 1.
n(CaCO₃) = n(CaO).
n(CaCO₃) = 0,89 mol.
m(CaCO₃) = m(CaCO₃) · M(CaCO₃).
m(CaCO₃) = 0,89 mol · 100 g/mol
m(CaCO₃) = 89 g.
n - amount of substance.
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When 0.50 L of a 12 M solution is diluted to 1.0 L, what is the resulting molarity?
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Answer:

The resulting molarity is 6M.

Explanation:

A dilution consists of the decrease of concentration of a substance in a solution (the higher the volume of the solvent, the lower the concentration).

We use the formula for dilutions:

C1 x V1 = C2 x V2

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3 years ago
How many grams of fluorine are contained in 8 molecules of boron trifluoride?
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<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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