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:
heat energy is released into the surrounding
Answer:
C. Yes, because they have a definite composition.
Explanation:
<h2><u><em>
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Answer:Compact bone, also called cortical bone, dense bone in which the bony matrix is solidly filled with organic ground substance and inorganic salts, leaving only tiny spaces (lacunae) that contain the osteocytes, or bone cells. Compact bone makes up 80 percent of the human skeleton; the remainder is cancellous bone, which has a spongelike appearance with numerous large spaces and is found in the marrow space (medullary cavity) of a bone. Both types are found in most bones. Compact bone forms a shell around cancellous bone and is the primary component of the long bones of the arm and leg and other bones, where its greater strength and rigidity are needed.
Explanation:
