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:
ethier a dessert or a plains
though plains can get rain in the summer
from me living in both
it seems more like a dessert
Answer:
40%
Explanation:
Calculate the percent of the mass that is carbon:
Explanation:
a) In 1 mole of methane there are 4 moles of hydrogen atom
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of methane = 4 × 1 g = 4 g
b) In 1 mole of chloroform there are 1 mole of hydrogen atom
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of methane = 1× 1 g = 1 g
c) In 1 mole of 
there are 10 moles of hydrogen atom
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of = 10 × 1 g = 10 g
d)In 1 mole of 
there are 12 moles of hydrogen atom.
Atomic mass of 1 mole of hydrogen atom = 1 g
Mass of hydrogen in 1 mole of = 12 × 1 g = 12 g
1: viewing any chemical reaction in a laboratory
2: dangerous to look at when it burns & used in photography, fireworks, and flares
3: the product
