Explanation:
1)
Mass of NaOH = m
MOlar mass of NaOH = 40 g/mol
Volume of NaOH solution = 1.00 L
Molarity of the solution= 1.00 M
A student can prepare the solution by dissolving the 40. grams of NaOH in is small volume of water and making that whole volume of solution to volume of 1 L.
Upto two significant figures mass should be determined.
2)
(dilution equation)
Molarity of the NaOH solution = 
Volume of the solution = 
Molarity of the NaOH solution after dilution = 
Volume of NaOH solution after dilution= 
A student can prepare NaOH solution of 1.00 M by diluting the 0.500 L of 2.00 M solution of NaOH with water to 1.00 L volume.
Upto three significant figures volume should be determined.
Answer:
here
Explanation:
0.000141 to kilowatt-hours. hope this helped
Answer:
588.2 mL
Explanation:
- FeSO₄(aq) + 2KOH(aq) → Fe(OH)₂(s) + K₂SO₄(aq)
First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):
- moles = molarity * volume
- 187 mL * 0.692 M = 129.404 mmol Fe⁺²
Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:
- 129.404 mmol Fe⁺² *
= 258.808 mmol KOH
Finally we<u> calculate the required volume of KOH solution</u>, using <em>the given concentration and the calculated moles</em>:
- volume = moles / molarity
- 258.808 mmol KOH / 0.440 M = 588.2 mL
To find the mass you need to find the weight of a mol of the molecules by adding up the atomic mass.
N = 14.007 g/mol
H = 1.008 g/mol
S = 32.065 g/mol
O = 16 g/mol
2(14.007) + 8(1.008) + 32.065 + 4(16) = 132.143 g/mol
Now you know how much an entire mol weight you multiply it by how much you actually have
0.00456 * 132.143 = 0.603 g
Answer:
0.758 V.
Explanation:
Hello!
In this case, case when we include the effect of concentration on an electrochemical cell, we need to consider the Nerst equation at 25 °C:
Whereas n stands for the number of moles of transferred electrons and Q the reaction quotient relating the concentration of the oxidized species over the concentration of the reduced species. In such a way, we can write the undergoing half-reactions in the cell, considering the iron's one is reversed because it has the most positive standard potential so it tends to reduction:
It means that the concentration of the oxidized species is 0.002 M (that of nickel), that of the reduced species is 0.40 M and there are two moles of transferred electrons; therefore, the generated potential turns out:
Beat regards!
