0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary to reach stoichiometric quantities with cacl2.
<h3>Explanation:</h3>
Based on the reaction
CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃
1 mole of CaCl₂ reacts per mole of Na₂CO₃
we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g
- We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
- These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
- Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:
<h3>
Moles CaCl₂.2H₂O:</h3>
1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂
Moles Na₂CO₃:
0.0102 moles Na₂CO₃
Mass Na₂CO₃:
0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present
Therefore, we can conclude that 0.0102 moles Na₂CO₃ is necessary.to reach stoichiometric quantities with cacl2.
To learn more about stoichiometric quantities visit:
<h3>
brainly.com/question/28174111</h3>
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Answer:
Inferring is when a scientist uses reasoning to explain or interpret the things they observe
Answer:
Enthalpy of formation = -947.68KJ/mol
Explanation:
Enthalpy of formation is the heat change when one mole of a substance is formed from its element in its standard states and in standard conditions of temperature and pressure. it may be positive or negative, if positive, it is an endothermic reaction where the heat content of the product is greater than that of the reactants, and if negative, it is exothermic reaction - where the heat content of the reactants is greater than the products. the enthalpy of formation is measured in KiloJoule/Moles (KJ/Mole).
From the value of the enthalpy of formation of NaHCO3, it shows that the reaction is exothermic, that is the formation of NaHCO3 from its constituents elements. As such, the heat content of the reactants is greater than the products.
The step by step explanation is shown in the attachment.
Answer: It is an unsaturated solution
Explanation: This is because it has more solute than a normal solution.
