Answer:
Benzene shows that it is actually unsaturated because it adds hydrogen or chlorine, although only when allowed to react under very vigorous conditions (higher temperature or pressure) compared to those required for alkenes and alkynes.
Explanation:
Answer:
me no speak espanol fella
0.4575 g of saccharine is present in the ten tablets of saccharine dissolved in water.
From the information in the question, the sulfur in saccharin (C7H5NO3S) was completely converted to sulfate ion (SO4^2-). This ion was now reacted with excess barium chloride to form barium sulfate BaSO4. This later reaction occurs as follows;
SO4^2-(aq) + Ba^2+(aq) --------> BaSO4(s)
Number of moles of BaSO4 obtained = 0.5240 g/233 g/mol
= 0.0025 moles
Since the reaction is 1:1, 0.0025 moles of sulfate ions from saccharine reacted.
Molar mass of saccharine = 7(12) + 5(1) + 14 + 3(16) + 32
= 84 + 5 + 14 + 48 + 32 = 183 g/mol
Mass of saccharine in the tablet = 0.0025 moles × 183 g/mol
= 0.4575 g
Learn more: brainly.com/question/1527403
Answer:
The correct answer is 0.033 M
Explanation:
We have a solution of NaClO with a concentration of 5%w/w:
5% w/w= 5 g NaClO/100 g solution
The first dilution is 10 ml of solution in 100 ml. That is a 1/10 dilution (10ml/100 ml= 1/10). That means we are diluting 10 times the solution. We can calculate the resulting concentration after this first dilution as follows:
5%w/w x 10 ml/100 ml = 5% w/w/10= 0.5%w/w
Then, we take 6 ml of 0.5% w/w solution and we add 6 ml of dye in a reaction vessel. The total volume of the solution in the reaction vessel is 6 ml + 6 ml= 12 ml, and we are diluting twice the solution because 6 ml/12 ml= 1/2. We can calculate the resulting concentration of the solution after this second dilution as follows:
0.5% w/w x 6 ml/12 ml= 0.5% w/w/2= 0.25%w/w
Finally, we need to convert the concentration from %w/w to M (mol solution/1L solution). For this, we assume a density of the solution close to the density of water (1.00 g/ml) and we use the molecular weight of NaClO (74.44 g/mol):
0.25 g NaClO/100 g solution x 1 mol NaClO/74.44 g x 1.00 g solution/1 ml x 100 ml/0.1 L= 0.033 mol/L
= 0.033 M
