Answer:
The answer to your question is
4C₇H₁₇ + 45 O₂ ⇒ 28 CO₂ + 34H₂O
Explanation:
Write the equation
C₇H₁₇ + O₂ ⇒ CO₂ + H₂O
Process
1.- Check if the equation is balanced
Reactants Element Products
7 C 1
17 H 2
2 O 3
As the number of reactants and products is different, we conclude that the reaction is unbalanced.
2.- Write a coefficient "7" to CO₂ and a coefficient of 17/2 to H₂O
C₇H₁₇ + O₂ ⇒ 7CO₂ + 
H₂O
Reactants Element Products
7 C 7
17 H 17
2 O 51/2
3.- Write a coefficient of 45/2 to the O₂, and multiply all the equation by 2.
4C₇H₁₇ + 45 O₂ ⇒ 28 CO₂ + 34H₂O
Reactants Element Products
28 C 28
68 H 68
90 O 90
Answer:
Moon has to be in-between the Earth and the Sun.
2. Moon's umbra should sweep your place.
3. Latitude and longitude of your place should be within the befitting limits.
Answer:
11
Explanation:
An atomic number of 11 means this atom will have 11 protons. A mass number of 23 means 23 - 11 this atom will have 12 neutrons. Since this atom is neutral the positive protons must be equal to the negative electrons. This atom will have 11 electrons
The Answer to your question would be A
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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