Answer:
6.7970 g
Explanation:
Considering the Henderson- Hasselbalch equation for the calculation of the pOH of the basic buffer solution as:
pOH = pK + log[acid] / [base]
Where K is the dissociation constant of the base.
Base dissociation constant of the ammonia = 1.8×10⁻⁵
pK = - log (Kb) = - log (1.8×10⁻⁵) = 4.75
Given concentration of base = [base] = 0.273 M
pH = 10.150
pOH = 14 - pH = 14 - 10.150 = 3.85
So,
3.85 = 4.75 + log[acid]/0.273
[Acid] = 0.0347 M
Given that Volume = 2 L
So, Moles = Molarity × Volume
Moles = 0.0347 × 2 = 0.0694 moles
Molar mass of ammonium bromide = 97.94 g/mol
Mass = Moles × Molar mass = (0.00775 × 97.94) g = 6.7970 g
Answer:
Explanation:
Method 1 proportion
1 mole of chromium is 52 grams
11.9 moles = x grams
1/11.9 = 52/x Cross multiply
x = 11.9 * 52
x = 618.8 grams
Now I have used an approximate mass for Chromium. The answer you get here is expected to reflect the weigth given on your periodic table Use that to get your answer. You should give a number very close to mine. Round to 3 places as in 619.
Method Two Formula
mols = given mass / molecular mass
11.9 = given mass / 51.9961 Multiply both sides by 51.9961
11.9 *51.9961 = given mass
given mass = 618.75
given mass = 619
Answer:
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Explanation:
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Answer:
O2 is the limiting reactant.
Explanation:
Hello there!
In this case, according to the given information about this reaction, one could identify the limiting reactant by performing a mole ratio of H2S to SO2 and O2 to SO2:
Thus, since 17.1 moles of O2 yields fewer moles than 14.3 moles of H2S, we infer the former is the limiting reactant.
Regards!