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lora16 [44]
2 years ago
15

Why it is difficult to accurately represent ammonia two-dimensionally?

Chemistry
1 answer:
Svetlanka [38]2 years ago
8 0
According to the valence shell electron pair repulsion (VSPER) theory, an ammonia molecule <span> has a </span>trigonal pyramidal<span> shape with an experimental bond angle measure of 106.7 degrees. This is why it is difficult to accurately represent ammonia two-dimensionally because the molecular structure entails a 3-D projection with angles in it unlike the linear structure.</span>
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If 6.00 g of CaCl2 • 2 H2O and 5.50 g of Na2CO3 are allowed to react in aqueous solution, what mass of CaCO3 will be produced? P
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Answer:

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of  CaCl₂ .2H₂O  × 100 g CaCO₃ /  1 mole CaCO₃ = 4 g

5.50 g Na₂CO₃   /1 × 1 Na₂CO₃  / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃  × 100 g CaCO₃ /  1 mole CaCO₃ = 5 g

Explanation:

Given data:

Mass of CaCl₂.2H₂O = 6.00 g

Mass of Na₂CO₃ = 5.50 g

Mass of CaCO₃ produced = ?

Solution:

Number of moles of CaCl₂.2H₂O.

Number of moles = mass/ molar mass

Number of moles = 6.00 g/ 147 g/ mol

Number of moles = 0.04 mol

Number of moles of Na₂CO₃:

Number of moles = mass/ molar mass

Number of moles = 5.50 g/ 106 g/ mol

Number of moles = 0.05 mol

Chemical equation:

CaCl₂  +  Na₂CO₃   →   CaCO₃ + 2NaCl

Now we will compare the moles of CaCO₃  with Na₂CO₃  and CaCl₂ through balanced chemical equation .

                      CaCl₂              :               CaCO₃

                             1                :                1

                       0.04               :            0.04

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.04 mol× 100 g/mol

Mass = 4 g

6.00 g CaCl₂ .2H₂O /1 × 1 mol CaCl₂ .2H₂O / 147 g CaCl₂ .2H₂O × 0.04 mol CaCO₃/ 0.04 mol of  CaCl₂ .2H₂O  × 100 g CaCO₃ /  1 mole CaCO₃ = 4 g

                     Na₂CO₃            :            CaCO₃

                          1                   :                1

                       0.05               :            0.05

Mass of CaCO₃:

Mass = number of moles × molar mass

Mass = 0.05 mol× 100 g/mol

Mass = 5 g

5.50 g Na₂CO₃   /1 × 1 Na₂CO₃  / 106 g Na₂CO₃ × 0.05 mol CaCO₃/ 0.05 mol of Na₂CO₃  × 100 g CaCO₃ /  1 mole CaCO₃ = 5 g

                     

5 0
3 years ago
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