Answer:
Metals are located on the left side of the periodic table and are generally shiny, malleable, ductile, and good conductors.
Explanation:
In the crystallization process the solid compound is dissolved in the solvent at elevated temperature and the crystallize product obtained by slow cooling of the solution. Here the solubility of acetanilide at 100°C is 1g per 20mL of water. Thus to dissolve 500mg of acetanilide at high temperature that is 100°C we need 10mL of water.
Now at 25°C after the re-crystallization there will be some amount of dissolve acetanilide. Which can be calculated as- 185mL of water is needed to dissolve 1g or 1000mg of acetanilide at 25°C. Thus in 10mL of water there will be
gmg of acetanilide.
Answer:
-The other substances that give a positive test with AgNO3 are other chlorides present, iodides and bromide.
-It is reasonable to exclude iodides and bromides but it is not reasonable to exclude other chlorides
Explanation:
In the qualitative determination of halogen ions, silver nitrate solution(AgNO3) is usually used. Now, various halide ions will give various colours of precipitate when mixed with with silver nitrate. For example, chlorides(Cl-) normally yield a white precipitate, bromides(Br-) normally yield a cream precipitate while iodides (I-) normally yield a yellow precipitate. Thus, all these ions or some of them may be present in the system.
With that being said, if other chlorides are present, they will also yield a white precipitate just like KCl leading to a false positive test for KCl. However, since other halogen ions yield precipitates of different colours, they don't lead to a false test for KCl. Thus, we can exclude other halides from the tendency to give us a false positive test for KCl but not other chlorides.
<h3>Answer:</h3>
0.64 Moles of Propane
<h3>Explanation:</h3>
Data:
Moles of Carbon = 1.5 mol
Conversion factor = 7 mol C produces = 3 mol of Propane
Solution:
As we know,
7 moles of Carbon produces = 3 moles of Propane
Then,
1.5 moles of Carbon will produce = X moles of Propane
Solving for X,
X = (1.5 moles × 3 moles) ÷ 7 moles
X = 0.6428571 moles of Propane
Or rounded to two significant figures,
X = 0.64 Moles of Propane
<em><u>the</u></em><em><u> </u></em><em><u>number</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>neutrons</u></em><em><u> </u></em><em><u>i</u></em><em><u>n</u></em><em><u> </u></em><em><u>aluminium</u></em><em><u> </u></em><em><u>is</u></em><em><u> </u></em><em><u>1</u></em><em><u>4</u></em>