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deff fn [24]
2 years ago
14

Calculate the percentage of sodium and carbon and oxygen in sodium carbonate​

Chemistry
1 answer:
Fantom [35]2 years ago
8 0
Percentage of Sodium [Na] in Sodium Carbonate [Na2CO3] = 46/106 x 100 = 43.40%. Percentage of Carbon [C] in Sodium Carbonate [Na2CO3] = 12/106 x 100 = 11.32%. Percentage of Oxygen [O] in Sodium Carbonate [Na2CO3] = 48/106 x 100 = 45.28% .....
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Answer:

The percent by mass of water in this crystal is:

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Explanation:

This exercise can be easily solved using a simple rule of three where the initial weight of the hydrated crystal (6,235 g) is taken into account as 100% of the mass, and the percentage to which the mass of 4.90 g corresponds (after getting warm). First, the values and unknown variable are established:

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And the value of the variable X is found:

  • X = (4.90 g * 100%) / 6,235 g
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The calculated value is not yet the percentage of the water, since the water after heating the glass has evaporated, therefore, the remaining percentage must be taken, which can be calculated by subtraction:

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Answer:

                       0.7644 moles of H₂O

Explanation:

                    The balance chemical equation is as follow;

                                      2 H₂O₂ → 2 H₂O + O₂

To solve this problem we will first calculate the moles of H₂O₂ as,

                           Moles  =  Mass / M/Mass

                           Moles  =  26.0 g / 34.01 g/mol

                           Moles  =  0.7644 mol

Secondly,

According to equation,

                      2 moles of H₂O₂ produces  =  2 moles of H₂O

Hence,

                0.7644 mol of H₂O₂ will produce  =  X moles of H₂O

Solving for X,

                        X  =  2 mol × 0.7644 mol / 2 mol

                       X =  0.7644 moles of H₂O

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