The question is incomplete, the complete question is; Describe the preparation of 2.00 L of 0.108 M BaCl2 from BaCl2.2H2O
Answer:
See explanation
Explanation:
First we must know the molar mass of the compound = 244.3 g/mol
Now we must calculate the number of moles of BaCl2.2H2O in the solution.
number of moles = concentration * volume = 0.108 M * 2.00 L = 0.216 moles
Now the mass of solid that we need to take can be obtained from;
number of moles = mass/molar mass
mass = number of moles * molar mass
mass = 0.216 moles * 244.3 g/mol
Mass = 52.8 g
Hence, we must weigh 52.8 g of BaCl2.2H2O accurately in a balance and dissolve it in 2 L of water to give 0.108 M solution of BaCl2.2H2O.
Iodine value is a measure of the degree of unsaturation in fats and oils. It is essentially the number of grams of iodine consumed by 100 g of fat. If the iodine number is in the range of 0-70 then it is a fat, any value above 70 is considered an oil.
Formula:
Iodine number = (ml of 0.1 N Thiosulphate blank- ml of 0.1N thiosulphate test) * 12.7 *100/1000* wt of sample
vol of thiosulphate required to titrate test sample (given oil) = 1 ml
wt of sample = 0.2 g
Information on the volume of thiosulphate required to titrate the blank solution is essential for calculation.
Iodine number = (X-1.0) * 12.7 * 100/1000* 0.2 = (X-1.0)*6.35
Given -
T = 450K
∆H = 3000J
∆S = 4J/K
Using formula
∆G = ∆H - T∆S
∆G = 3000- 4*450
∆G = 1200 Joule
∆G is positive so the reaction is non spontaneous!
➢225 mL of solution containing<span> 0.486 </span>mole<span> of solute </span>mol<span> HCl. Use </span>molarity<span> as a conversion factor.</span>Mol<span> HCl = 2.5 L soln x 0.10 </span>mol<span> HCl by diluting 25.0 mL of 6.00 M HCl to a </span>total volume<span> of 50.0 Note: Vc and Vd do not </span>have<span> to be ➢The mass of MgCl2 in a </span>2.75 mole<span> sample 3.447 g of </span>CO2<span> and 1.647 g of H2O.</span>
