Answer:
1.096g
Explanation:
You must know the atomic mass of Hydrogen, Fluorine, and Sodium before you can start:
Hydrogen: 1.008g/mol
Fluorine: 18.99g/mol
Sodium: 22.98g/mol
Next, find the composition percentage of NaF
22.98 + 18.99 = 41.97
Fluorine is 18.99/41.97 =45.25%
Sodium is 100-45.25 = 54.75%
Ultimately we want to know about HF so find how much F is in 2.3g: 2.3 * 0.4525 = 1.041g
Find comp. percentage of HF
18.99+1.008 = 19.998; H/total F/total
Hydrogen 5.041%
Fluorine 94.959%
Laws of conservation of say we have 1.041g of fluorine in our HF. We know 1.041 is 94.959% of the mass of HF so do some simple math to find the remaining: 1.041/0.94959 = 1.096g
Explanation:
this equation is balanced
if you look at it carefully
k=1
f=1
o=2
we do not have any opposing element
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%
Answer: It should a accepter
Explanation: because water is often used as solvent for solutes...there are exceptions
F because it NEVER forms any cations in chemical reactions