<span>1.15x10^24 molecules of hypothetical substance b
Making the assumption that each molecule in hypothetical substance a reacts to produce a single molecule of hypothetical substance b, then the number of molecules of substance b will be the number of moles of substance a multiplied by avogadro's number. So
Moles hypothetical substance a = 29.9 g / 15.7 g/mol = 1.904458599 moles
This means that we should also have 1.904458599 moles of hypothetical substance b. And to get the number of atoms, multiply by 6.0221409x10^23, so:
1.904458599 * 6.0221409x10^23 = 1.146892x10^24 molecules.
Rounding to 3 significant figures gives 1.15x10^24</span>
Answer:
Cupric oxide, or copper (II) oxide,
Explanation:
Correct me if im wrong tnx:<
Answer:
37.1g are produced
Explanation:
The combustion of C₈H₁₈ is:
C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O
<em>Where 1 mole of C₈H₁₈ produce 8 moles of CO₂</em>
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To find the mass of CO₂ that is produced we need to convert the mass of C₈H₁₈ with molar mass. Then, with the chemical equation, we can find the moles of CO₂ and its mass, as follows:
<em>Moles C₈H₁₈ -Molar mass: 114.2g/mol-</em>
12g C₈H₁₈ * (1mol / 114.2g) = 0.105 moles of C₈H₁₈
<em>Moles CO₂:</em>
As 1 mole of C₈H₁₈ produce 8 moles of CO₂, 0.105 moles of C₈H₁₈ produce:
0.105 moles of C₈H₁₈ * (8moles CO₂ / 1mole C₈H₁₈) = 0.84 moles of CO₂
<em>Mass CO₂ -Molar mass: 44.01g/mol-:</em>
0.84 moles of CO₂ * (44.01g / mol) = 37.0g of CO₂ ≈
<h3>37.1g are produced</h3>
The molar concentration of a sucrose solution prepared by dissolving 350.25 g of sucrose is enough deionized water to yield a final solution volume of 500.00 mL is 2.048 M.
We must first obtain the number of moles of sucrose in the solution as follows;
Number of moles = mass/ molar
Mass of sucrose = 350.25 g
Molar mass of sucrose = 342 g/mol
Number of moles of sucrose= 350.25 g / 342 g/mol
= 1.024 moles
Recall that;
Number of moles = concentration × volume
concentration = Number of moles/volume
volume of solution = 500.00 mL or 0.5 L
concentration = 1.024 moles/0.5 L
concentration = 2.048 M
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