Answer:
Grey precipitate implies the presence of silver ions
Yellow precipitate implies the presence of lead II ions
Explanation:
Qualitative analysis provides us a quick method of identifying ions present in a sample by chemical reactions involving simple reagents. Precipitates having a unique colour is formed. The identity of ions in the sample is deduced from the colour of precipitate obtained when particular reagents are added.
In the question, a precipitate containing silver ions upon standing turn into grey colour. Similarly, lead II ions give a yellow precipitate.
Answer:
V2= 1.03L
Explanation:
Start off with what you are given.
V^1: 1.00L
T^1: 23°C
V^2?
T^2: 33°C
If you know your gas laws, you have to utilise a certain gas law called Charles' Law:
V^1/T^1 = V^2/T^2
Remember to convert Celsius values to Kelvin whenever you are dealing with gas problems. This can be done by adding 273 to whatever value in Celsius you have.
(23+273 = 296) (33+273 = 306)
Multiply crisscross
1.00/296= V^2/306
296V^2 = 306
Dividing both sides by 296 to isolate V2, we get
306/296 = 1.0337837837837837837837837837838
V2= 1.03L
I think it’s d
If it’s wrong then I’m sorry
Answer:- C. Hafnium.
Solution:- Mass of the sample is 46.0 g and it's volume is 
.
From mass and volume, we can calculate it's density using the formula:
On the basis of the density, this substance could either be mercury or hafnium. Since the substance is a solid at room temperature where as mercury is liquid. So, it can't be mercury.
The right choice is C) Hafnium.
The type of bond most likely formed in nitrous oxide would be a covalent bond..
