Answer : The Lewis-dot structure of
is shown below.
Explanation :
Lewis-dot structure : It shows the bonding between the atoms of a molecule and it also shows the unpaired electrons present in the molecule.
In the Lewis-dot structure the valance electrons are shown by 'dot'.
The given molecule is, 
As we know that hydrogen has '1' valence electron and nitrogen has '5' valence electrons.
Therefore, the total number of valence electrons in
= 5 + 3(1) = 8
According to Lewis-dot structure, there are 6 number of bonding electrons and 2 number of non-bonding electrons.
Now we have to determine the formal charge for each atom.
Formula for formal charge :





Hence, the Lewis-dot structure of
is shown below.
Molarity = (Mass/ molar mass) x (1/ volume of solution in Litres)
Mass = Molarity x molar mass x volume of solution in Litres
Molarity of Tris = 100 mM = 0.1 M
volume of Tris sol. = 100 mL = 0.1 L
molar mass of Tris = 121.1 g/mol
Hence,
mass of Tris = Molarity of Tris x molar mass ofTris x volume of Tris solution
= 0.1 M x 121.1 g/mol x 0.1 L
= 1.211 g
mass of Tris = 1.211 g
True because it doesn’t count as a full number
Answer:
repel
Explanation:
When it comes to electrical forces, "opposites charges attract" while "like charges repel."
There are primarily two types of charges: positive charge and negative charge. The forces they exert upon each other will depend on their charges. The<u> positive charge has an </u><em><u>attractive force</u></em><u> to a negative charge.</u> On the contrary,<u> it has a</u><em><u> repulsive force</u></em><u> to the same positive charge</u>. Thus, it will repel each other.
So this means that <em>opposite charges will draw closer together</em> while<em> like charges will move apart from each other.</em>
To present his data on the <span>solubility of three different salts in water at 22°C, bar graph should be used since there are different salts and the only variable is the type of salt used. Line graph and scatter plot use two coordinates or variables and is common to comparing data using the same sample while using histogram to find out data distribution is irrelevant.</span>