You will feel way better because in our class we learned that lukewarm baths are good for fevers and warm washcloths :)
Answer:
Percent Yield Fe = 82.5%
Explanation:
The actual yield is the value produced after an experiment is conducted. The theoretical yield is the value calculated using the balanced chemical equation and atomic/molar masses.
To find the percent yield of iron (Fe), you need to (1) convert grams Al to moles Al (via atomic mass), then (2) convert moles Al to moles Fe (via mole-to-mole ratio from equation coefficients), then (3) convert moles Fe to grams Fe (via atomic mass), and then (4) calculate the percent yield. It is important to arrange the ratios in a way that allows for the cancellation of units. The final answer should have 3 sig figs to reflect the sig figs of the given values.
Atomic Mass (Mg): 24.305 g/mol
Atomic Mass (Fe): 55.845 g/mol
3 Mg + 2 FeCl₃ -----> 2 Fe + 3 MgCl₂
20.5 g Mg 1 mole 2 moles Fe 55.845 g
----------------- x ----------------- x ---------------------- x ----------------- =
24.305 g 3 moles Mg 1 mole
= 31.4 g Fe
Actual Yield
Percent Yield = ---------------------------------- x 100%
Theoretical Yield
25.9 g Fe
Percent Yield = -------------------- x 100%
31.4 g Fe
Percent Yield = 82.5%
The answer is going to be C) because
A)We know that she put all three boxes in the window and let them sat their for a hour
B)We know that the boxes are the same size
C)She used three different materials,and she wants to measure the warmth of each box. To see which one is the warmest
D)She never said anything about mass
Hopes This Helps:)
I want to say A only... Hope I helped!!
Answer:
37.8 L OF CARBON MONOXIDE IS REQUIRED TO PRODUCE 18.9 L OF NITROGEN.
Explanation:
Equation for the reaction:
2 CO + 2 NO ------> N2 + 2 CO2
2 moles of carbon monoxide reacts with 2 moles of NO to form 1 mole of nitrogen
At standard temperature and pressure, 1 mole of a gas contains 22.4 dm3 volume.
So therefore, we can say:
2 * 22.4 L of CO produces 22.4 L of N2
44.8 L of CO produces 22.4 L of N2
Since, 18.9 L of Nitrogen is produced, the volume of CO needed is:
44.8 L of CO = 22.4 L of N
x L = 18.9 L
x L = 18.9 * 44.8 / 22.4
x L = 18.9 * 2
x = 37.8 L
The volume of Carbon monoxide required to produce 18.9 L of N2 is 37.8 L
