Answer:
48
Explanation:
because you add 6 and 6 and 12 to get it
Answer:
3 electrons
Explanation:
aluminum : [Ne]3s23p1 [ N e ] 3 s 2 3 p 1 . It loses 3 electrons from 3s and 3p orbitals and attains the noble gas configuration of Neon.
For the first question, salt is soluble while sand is insoluble or not dissolvable in water. The salt should have vanished or melted, but the sand stayed noticeable or visible, making a dark brown solution probably with some sand particles caught on the walls of the container when the boiling water was put in to the mixture of salt and sand. The solubility of a chemical can be disturbed by temperature, and in the case of salt in water, the hot temperature of the boiling water enhanced the salt's capability to melt in it.
For the second question, the melted or dissolved salt should have easily made its way through the filter paper and into the second container, while the undissolved and muddy sand particles is caught on the filter paper. The size of the pores of the filter paper didn’t change. On the contrary, the size of the salt became smaller because it has been dissolved which is also the reason why it was able to go through the filter paper, while the size of the sand may have doubled or even tripled which made it harder to pass through.
Answer:
5200 ppm
Explanation:
As per the definition, parts per million of a contaminant is a measure of the amount of mass of contaminant present per million amount of the solution. It is denoted by ppm.
Given in the question,
Water = 250 ml = 250 g
Lead = 1.30 g
So,
ppm of Lead = 
= 
= 5200 ppm
So, as calculated above, there is 5200 ppm of lead present in 250 ml of water.
Answer: 400K
Explanation:
Given that,
Original volume of balloon V1 = 3.0L
Original temperature of balloonT1 = 27°C
Convert the temperature in Celsius to Kelvin
(27°C + 273 = 300K)
New volume of balloon V2 = 4.0L
New temperature of balloon T2 = ?
Since volume and temperature are given while pressure is constant, apply the formula for Charle's law
V1/T1 = V2/T2
3.0L/300K = 4.0L/T2
To get the value of T2, cross multiply
3.0L x T2 = 4.0L x 300K
3.0LT2 = 1200LK
Divide both sides by 3.0L
3.0LT2/3.0L = 1200LK/3.0L
T2 = 400K
Thus, at a temperature of 400 Kelvin, the balloon would have a volume of 4.0L.
