Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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The electronegativity of nitrogen (N) is 3.0, while the electronegativity of hydrogen (H) is 2.1. As it can be seen that nitrogen (N) is more electronegative than that of hydrogen (H),
So electron pairs are attracted towards nitrogen and thus it carries a partial negative charge and hydrogen carries a partial positive charge. The image of electron distribution is attached as follows.
Thus NH₃ is a polar molecule .
Yes, you should memorize them. While it is easy to predict them by looking at a periodic table, you should still have a strong grasp of simpler chemistry such as this before more challenging topics are taught.
The balanced equation is attached in the image below. The coefficients are 2, 2, blank.
To get the number of gold atoms, you have to divide the mass of the gold by the mass of the gold atom. It follows this simple equation

.
Let x be the number of gold atoms. Plug in the values to a calculator.
x =

Both have the same units so the unit gram(g) can be cancelled.
x then would be equal to 1.53x10^22. So there are 1.53x10^22 atoms of gold in 5 g of gold