The molarity of the lake water is 0.00001 M and the pH of lake water is 5.
The lake water is acidic.
Explanation:
Data given:
molarity of base solution Mbase = 0.1 M
volume of the base solution Vbase = 0.1 ml or 0.0001 litre
volume of lake water Vlake = 1000ml or 1 litre
molarity of the lake water, Mlake = ?
Using the formula for titration:
Mbase X Vbase = Mlake X
Mlake = 
Putting the values in the equation:
Mlake = 
Mlake = 0.00001 M
The pH of the lake water will be calculated by using the following formula:
pH = - 
[
]
pH = - [ 0.00001]
pH = 5
Answer:
The correct answer is 4.58 grams.
Explanation:
Based on the Faraday's law of electrolysis, at the time of electrolysis, the amount of deposited substance is directly equivalent to the concentration of the flow of charge all through the solution. If current, I, is passed for time, t, seconds and w is the concentration of the substance deposited, then w is directly proportional to I*t or w = zIt (Here z refers to the electrochemical equivalent or the amount deposited when 1 C is passed).
For the reaction, n * 96500 C = molar mass
1C = molar mass/n*96500 = Equivalent wt / 96500
w = Equivalent wt / 96500 * I * t
In the given reaction,
Pb + PbO2 + 2HSO4- + 2H+ → 2PbSO4 + 2H2O, n = 2, the current or I drawn is 350 A, for time, t 12.2 seconds.
Now putting the values in the equation we get,
w = 207.19 / 2 * 96500 * 350 * 12.2 ( The molecular weight of Pb is 207.19 and the equivalent weight of Pb is 207.19 / 2)
w = 4.58 gm.
Answer:
A
Explanation:
If the Picture is a Desert showing reptiles like lizards and has cacti then it would be Dry and Sandy
Structure along with ¹H-NMR is shown below.
Signal for Hₐ; Based on multiplicity of of the peak, a
Singlet peak (the only singlet peak present in spectrum) at
2.2 ppm was assigned to Hₐ. As the methyl group in not coupling with any other proton so it gave a singlet peak.
Signal for Hb; Based on multiplicity of of the peak, a
Quartet peak (the only quartet peak present in spectrum) at
2.4 ppm was assigned to Hb. As the methylene group in coupling with methyl group having three protons so it gave a Quartet peak.
Signal for Hc; Based on multiplicity of of the peak, a
Triplet peak (the only triplet peak present in spectrum) at
1.1 ppm was assigned to Hc. As the methyle group in coupling with methylene group having two protons so it gave a Triplet peak.
