the percent yield of the reaction is 100%.
The percent yield is calculated as the experimental yield divided by the theoretical yield x 100%:
% yield = actual yield / theoretical yield * 100%
% yield of a reaction in this case Rate
In this case, the molar mass of NaBr is 102.9 g / mol, as you know:
444 actual yield = 7.08 mol x 102.9 g / mol = 728.532 g
theoretical yield = 7.08 mol x 102.9 g / mol = 728.532 g
, Replaced by the definition of percent yield:
percent yield = 728.532 grams / 728.532 grams * 100%
percent yield = 100%
Finally, the percent yield of the reaction is 100%.
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FeBr3 is iron bromide. Also known as iron bromide. Iron bromide is an ionic compound in which iron is in a +3 oxidation state.
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Answer:
1.784 g
Explanation:
The equation of the reaction is;
NaOH(aq) + KHC8H4O4(aq) --------> KNaC8H4O4(aq) + H2O(l)
Number of moles of NaOH reacted = 17.47/1000 * 0.5000 M
Number of moles of NaOH reacted =8.735 * 10^-3 moles
From the reaction equation;
1 mole of NaOH reacted with 1 mole of KHC8H4O4
Hence, 8.735 * 10^-3 moles of NaOH reacts with 8.735 * 10^-3 moles of KHP.
So,
Mass of KHP reacted = 8.735 * 10^-3 moles * 204.2 g/mol = 1.784 g
Answer:
Answer: (b) F
Explanation:
Sodium has 1, magnesium has 2 and Aluminium has 3 electrons in its outermost shell whereas Fluorine has 7 electrons in its outermost shell hence Fluorine does not lose electrons easily.
The electronic configuration of fluorine is 2,7.
Fluorine is the ninth element with a total of 9 electrons.
The first two electrons will go in the 1s orbital.
The next 2 electrons for F go in the 2s orbital.
The remaining five electrons will go in the 2p orbital. Therefore the F electron configuration will be 1s22s22p5.
you have to show us the rest of it because we have no idea what your looking at. I'm sorry
Answer:
if an atom gains an electron, the ion has negetive charge
