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:
i am unique and have a specific number in the periodic table
POH of a 0.0072 M=-lg(0.0072) = 2.1426675
Answer:
The energy required to ionize the ground-state hydrogen atom is 2.18 x 10^-18 J or 13.6 eV.
Explanation:
To find the energy required to ionize ground-state hydrogen atom first we calculate the wavelength of photon required for this operation.
It is given by Bohr's Theory as:
1/λ = Rh (1/n1² - 1/n2²)
where,
λ = wavelength of photon
n1 = initial state = 1 (ground-state of hydrogen)
n2 = final state = ∞ (since, electron goes far away from atom after ionization)
Rh = Rhydberg's Constant = 1.097 x 10^7 /m
Therefore,
1/λ = (1.097 x 10^7 /m)(1/1² - 1/∞²)
λ = 9.115 x 10^-8 m = 91.15 nm
Now, for energy (E) we know that:
E = hc/λ
where,
h = Plank's Constant = 6.625 x 10^-34 J.s
c = speed of light = 3 x 10^8 m/s
Therefore,
E = (6.625 x 10^-34 J.s)(3 x 10^8 m/s)/(9.115 x 10^-8 m)
<u>E = 2.18 x 10^-18 J</u>
E = (2.18 x 10^-18 J)(1 eV/1.6 x 10^-19 J)
<u>E = 13.6 eV</u>
Answer:
B. double-replacement RXN
Explanation:
more specifically, this is a precipitation rxn.