Answer:
I don't really get the options but it favoures the reactant side.
Explanation:
Increasing pressure favours the side with fewer moles of gas while decreasing pressure favours the side with the more moles of gas. E.g
If there is 0 moles of gas particles in the reactant side and 1 mole of gas particle in the product side, increasing pressure favours the reactants while decreasing pressure favours the product side.
With the explanations I have made, I hope the question is now clear to you.
Answer: D is the correct answer
Explanation: When there is no insulator surrounding a metal wire then there is nothing to stop the electrons from flowing outside of the wire, This means that anything the wire comes in contact with could get shocked or burnt when there is a current flowing through the wire.
Answer:
Option D. KBr < KCl < NaCl
Explanation:
We'll begin by calculating the number of mole of each sample.
This can be obtained as follow:
For NaCl:
Mass = 1 g
Molar mass of NaCl = 23 + 35.5 = 58.5 g/mol
Mole of NaCl =?
Mole = mass /Molar mass
Mole of NaCl = 1/58.5
Mole of NaCl = 0.0171 mole
For Kbr:
Mass = 1 g
Molar mass of KBr = 39 + 80 = 119 g/mol
Mole of KBr =?
Mole = mass /Molar mass
Mole of KBr = 1/119
Mole of KBr = 0.0084 mole
For KCl:
Mass = 1 g
Molar mass of KCl = 39 + 35.5 = 74.5 g/mol
Mole of KCl =?
Mole = mass /Molar mass
Mole of KCl = 1/74.5
Mole of KCl = 0.0134 mole
Summary
Sample >>>>>>>> Number of mole
NaCl >>>>>>>>>> 0.0171
KBr >>>>>>>>>>> 0.0084
KCl >>>>>>>>>>> 0.0134
Arranging the number of mole of the sampl in increasing order, we have:
KBr < KCl < NaCl
The chemical symbol that represents the element silver is B. Ag
Answer: 0.52 L of 15 M 
will be used to prepare this amount of 0.52 M base.
Explanation:
But on diluting the number of moles remain same and thus we can use molarity equation.
(to be prepared)
where,
=concentration of stock solution = 15 M
= volume of stock solution = ?
= concentration of solution to be prepared = 0.52 M
= volume of solution to be prepared = 15 L
Thus 0.52 L of 15 M will be used to prepare this amount of 0.52 M base
