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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Kr 5s2 is the correct noble gas configuration for strontium
Answer:
combustion is fast and destructive it also involves a lot of movement of the atoms where erosion is a slow process that does not involve a lot of energy
Answer:
1. -4
2. +1
3. 0
4. +4
5. -2
6. +1
7. -2
reduced = H
oxidized = O
Explanation:
Know oxidation rules.
- Hope this helped! Please let me know if you would like to learn this. I could show you the rules and help you work through them.
20 minutes.
The sample would lose one half the quantity of francium in each half-life.
Thus a mass decrease by a factor of 16 would correspond to a period of four half-lives. It took 80 minutes for the sample to lose all these francium, therefore
minutes.
