To determine the amount of 6.0 M H2SO4 needed for the preparation, equate the number of moles of the 6.0 M and 2.5 M H2SO4 solution. This is done as follows
M1 x V1 = M2 x V2
Substituting the known variables,
(6.0 M) x V1 = (2.5 M) x (4.8 L)
Solving for V1 gives an answer of V1 = 2 L. Thus, to prepare the needed solution, dilute 2 L of 6.0 M H2SO4 solution with water until the volume reach 4.8 L.
Answer:
B. plants use the energy from sunlight to produce food.
hHope this helps! God loves you so much, that He died for you and would die a cruel death thousands of times for you. ✝️
B is the answer
hope this helps
2NBr₃ + 3NaOH = N₂ + 3HOBr + 3NaBr
40 mol 48 mol
NBr₃:NaOH = 2:3
40:48 = 2:2.4 = 2.5:3
NBr₃ is the excess reactant
Answer:
92.9%
Explanation:
You have been given the actual yield of the reaction. First, you need to find the theoretical yield of the reaction. To do this, you need to (1) convert grams Fe₂O₃ to moles Fe₂O₃ (via molar mass from periodic table values), then (2) convert moles Fe₂O₃ to moles Fe (via mole-to-mole ratio from reaction coefficients), and then (3) convert moles Fe to grams Fe (via molar mass).
Once you have found the theoretical yield, you need to use the percent yield equation to calculate the final answer. This number should have 3 sig figs to match the given values.
<u>(Step 1)</u>
Molar Mass (Fe₂O₃): 2(55.845 g/mol) + 3(15.998 g/mol)
Molar Mass (Fe₂O₃): 159.684 g/mol
1 Fe₂O₃(s) + 3 CO(g) ---> 2 Fe(s) + 3 CO₂(g)
Molar Mass (Fe): 55.845 g/mol
50.0 g Fe₂O₃ 1 mole 2 moles Fe 55.845 g
-------------------- x ------------------ x --------------------- x ---------------- = 35.0 g Fe
159.684 g 1 mole Fe₂O₃ 1 mole
<u>(Step 2)</u>
Actual Yield
Percent Yield = --------------------------- x 100%
Theoretical Yield
32.5 g Fe
Percent Yield = ---------------------- x 100% = 92.9%
35.0 g Fe
