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Aloiza [94]
3 years ago
13

16. The concentration of a solution of potassium hydroxide is determined by titration with nitric

Chemistry
1 answer:
____ [38]3 years ago
3 0

Answer:

M_{base}=0.709M

Explanation:

Hello,

In this case, since the reaction between potassium hydroxide and nitric acid is:

KOH+HNO_3\rightarrow KNO_3+H_2O

We can see a 1:1 mole ratio between the acid and base, therefore, for the titration analysis, we find the following equality at the equivalence point:

n_{acid}=n_{base}

That in terms of molarities and volumes is:

M_{acid}V_{acid}=M_{base}V_{base}

Thus, solving the molarity of the base (KOH), we obtain:

M_{base}=\frac{M_{acid}V_{acid}}{V_{base}} =\frac{0.498M*42.7mL}{30.0mL}\\ \\M_{base}=0.709M

Regards.

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Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho
Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3

c) The leftover is computed as follows:

m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0
3 years ago
A) How do you prepare %3 (w/v) Na2CO3 solution from Na2CO3⸱2H2O? (15p) Na2CO3 MW=106 g/mol
just olya [345]

Answer:

4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL

Explanation:

A 3%(w/v) solution contains 3g of solute (In this case, Na2CO3) in 100mL of solution.

Assuming we require 100mL of solution we must add 3g of Na2CO3. The reactant that is available is its dihydrate, with molar mass:

106g/mol + 2*MW H2O

106g/mol + 2*18g/mol = 142g/mol

That means the mass of Na2CO3.2H2O that must be added to prepare the solution is:

3g Na2CO3 * (142g/mol Na2CO3.2H2O / 106g/mol Na2CO3) =

<h3>4.02g of Na2CO3⸱2H2O must be added completing the volume of the solution to 100mL</h3>
8 0
3 years ago
6th grade help me pleaseee
chubhunter [2.5K]

Answer:

D hope this helps

Explanation:

6 0
3 years ago
Read 2 more answers
What percentage of energy used in the u.s is produced from fossil fuels
wel

Fossil Fuels Account for Lowest Share of U.S. Energy Consumption in More than a Century. Fossil fuels supplied about 80 percent of the energy consumed in the United States in 2017, the lowest share since 1902, according to a new report from the U.S. Department of Energy.

5 0
4 years ago
What is the volume of 14.0 g of nitrogen gas, N2, at a pressure of 2.00 atm and temperature of 27.0 degree celsius
Lelu [443]
P = 2.00 atm = 2026.50hPa
T = 27°C = 300K
m = 14.0g              |
Mr = 28g/mol         |      n = m/M = 14g/28g/mol = 0,5mol
R = 83.1 [hPa*L]/[mol*K]

pV = nRT

        nRT        0,5mol * 83.1[hPa*L]/[mol*K] * 300K
V = ---------- = ----------------------------------------------------------- = <u>6.15L</u>
           p                           2026.5hPa

Answer B.
3 0
3 years ago
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