2.187 is the answer to this question
Explanation:
Atomic number , protons and electrons have the same value / their value is same .
But for the neutron there is no specific technique. You have to remember the neutrons of every element
Answer:
True
Explanation:
It's true because the pH is a measure of how basic or acid a solution is. In an acidic medium, the pH scales goes from 0 to 7. While in a basic medium goes from 7 to 14. The lower the pH value of the most acid the solution is.
1. The expression pH = -log(molar concentration of hydronium) allow to calculate the pH of a solution.
2. On the other hand, the expression pOH = -log(molar concentration of hydroxide) allow to determine the pOH of a solution.
The values of pH and pOH always obey the following expression:
pH + pOH = 14
Thus if for instance the pH becomes smaller the pOH must become bigger in order to fulfill the equation. Which means that the concentration of hydronium ions is greater than the hydroxide concentration.
For example, in an acidic medium:
if pH= 3, pOH= 11
In this case the molar concentration of hydronium is 0,001M. And the molar concentration of hydroxide ions is just 0,00000000001M.
Answer:
2.03125g of acetylene
Explanation:
First thing's first, we have to write out the balanced chemical equation;
CaC2(s) + 2H2O(l) → Ca(OH)2(aq) + C2H2(g)
Water is in excess, so CAC2 is our limiting reactant. i.e it determines the amount of product that would be formed.
1 mol of CaC2 produces 1 mol of C2H2
In terms of mass;
Mass = Number of moles * Molar mass
where the molar mass of the elements are;
Ca = 40g/mol
C = 12g/mol
H = 1g/mol
CaC2 = 40+ (2*12) = 64g/mol
C2H2 =( 2 * 12) + ( 2 * 1) = 26g/mol
64g (1 * 64g/mol) of CaC2 produces 26g ( 1mol * 26g/mol) of C2H2
5g would produce x?
64 = 26
5 = x
Upon solving for x we have;
x = (5 * 26) / 64
x = 2.03125g
Answer:
p and d orbitals
Explanation:
A π bond forms when two orbitals overlap side-on.
The most common types are formed by the overlap of p orbitals (Fig. 1).
However, d orbitals can also overlap sideways with each other to form dπ-dπ bonds and with p orbitals to form dπ-pπ bonds (Fig.2).
