I think so that hypothesis is for isotopes and atomic mass.
6.0 will the pH at the neutralization point of 0.00812 m Ba(OH)2 be when titrated with HCl.
<h3><u>What are </u><u>
titration reactions ?</u></h3>
A titrant/titrator, a standard solution with fixed volume and concentration, is prepared as part of the titration procedure. When an endpoint or equivalence point is reached, the titrant is made to continue reacting with the analyte, and at that point, the amount of titrant consumed may be used to calculate the analyte's concentration. Alternately, titration is the application of the stoichiometry principle to determine an unknown solution's concentration.
The technique begins by adding a very little quantity of analyte to a beaker or Erlenmeyer flask. Under a calibrated burette or chemical pipetting syringe containing the titrant, a little quantity of an indicator (such as phenolphthalein) is inserted.
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Answer:
when the solution is heated and reaches its saturation point thus starts forming crystals thus crystals starts forming hence crystallisation takes place
Answer: IUPAC NOMENCLATURE
Explanation:
IUPAC stands for International Union of Pure and Applied Chemistry. They devised a systematic method for naming compounds in order to create a uniform global unambiguous system of nomenclature hence making it easier for researchers to share information more freely without the hindrance of reporting the same compound using different names in different parts of the world thus creating confusion in chemical literature.
To answer this question, we will use the following equation:
<span>ln(P2/P1) = (∆Hvap/R)*((1/T1) - (1/T2))
</span>
Now we examine the givens of the problem and transform to standard units if required:
<span>∆Hvap = 30.5 kJ/mol
</span>R is a constant = <span>8.314 x 10^-3 kJ K^-1 mol^-1
T1 </span><span>= 91 celcius = 91 + 273= 364 Kelvin
</span>T2 = 20 celcius = 20 + 273 = 293 k3lvin
P1 is the standard atmospheric pressure = 760 mmHg
P2 is the value to be calculated
Substitute with these values in the equation:
ln(P2/760) = (30.5 / 8.314 x 10^-3) x ((1 / 364) - (1 / 293))
ln(P2/760) = - 2.4662 (Take the exponential both sides to eliminate the ln)
P2 / 760 = e^(-2.4462) = 0.0866
P2 = 0.0866 x 760 = 65.816 mmHg
