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.
Answer:
oxygen should be the central atom in the middle
Explanation:
hydrogens always go on the outside
The correct answer to this question is that the length of 14 is it’s half Which would be 7
<span> By definition, </span>oxidation number<span> is the charge left on the given atom when all the bonding pairs (of electrons) are broken, so the oxidation number of Br will be +1</span>
Answer:
V₂ = 5.97 L
Explanation:
Given data:
Initial temperature = 9°C (9+273 = 282 K)
Initial volume of gas = 6.17 L
Final volume of gas = ?
Final temperature = standard = 273 K
Solution:
Formula:
The Charles Law will be apply to solve the given problem.
According to this law, 'the volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure'
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
V₂ = V₁T₂/T₁
V₂ = 6.17 L × 273K / 282 k
V₂ = 1684.41 L.K / 282 K
V₂ = 5.97 L
