Answer:
Because no two elements have the same set of energy levels, different elements emit different colors of light. Energy is released when electrons move from higher energy levels to lower ones (visible light).
Explanation:
Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
Answer:
1040%
Explanation:
To solve this question we must convert the mass of Iron to moles in order to find limiting reactant. With limiting reactant we can find the theoretical moles of hydrogen and theoretical mass:
Percent yield = Actual yield (5.40g) / Theoretical yield * 100
<em>Moles Fe -Molar mass: 55.845g/mol-:</em>
10.3g * (1mol / 55.845g) = 0.184 moles of Fe will react.
For a complete reaction of these moles there are necessaries:
0.184 moles Fe* ( 3 mol H2SO4 / 2 mol Fe) = 0.277 moles H2SO4.
As there are 14.8 moles of the acid, <em>Fe is limiting reasctant.</em>
The moles of H2 produced are:
0.184 moles Fe* ( 3 mol H2 / 2 mol Fe) = 0.277 moles H2
The mass is:
0.277 moles H2 * (2.016g/mol) = 0.558g H2
Percent yield is:
5.40g / 0.558g * 100 = 1040%
It is possible the experiment wasn't performed correctly
Answer:
103.9 g
Explanation:
First <u>we convert 54.0 g of propane (C₃H₈) into moles</u>, using its <em>molar mass</em>:
- 54.0 g ÷ 44 g/mol = 1.23 mol C₃H₈
Then we <u>convert 1.23 moles of C₃H₈ into moles of CO₂</u>, using the <em>stoichiometric coefficients</em>:
- 1.23 mol C₃H₈ *
= 3.69 mol CO₂
We <u>convert 3.69 moles of CO₂ into grams</u>, using its <em>molar mass</em>:
- 3.69 mol CO₂ * 44 g/mol = 162.36 g
And <u>apply the given yield</u>:
- 162.36 g * 64.0/100 = 103.9 g