Atomic # 30
Atomic mass- 65.409g
Symbol- Zn
protons = electrons,
so 30 protons,30 electrons and 35 neutrons.
The atomic mass is also found by adding the protons and neutrons.
It can be found in group 12,period 4. Zinc is a transition metal.
Zinc has an oxidation state of 2+.
Melting point- 420 C
Boiling point- 907 C
Typical compounds formed are: zinc oxide, zinc sulfide and zinc nitrate.
According to records, a German scientists named Andreas Margraf discovered the element in 1746. Even though he was recognized for the discovery, historians agree ancient people created statues with some amount of zinc.
Zinc can be found in united states, Canada and Australia.
Hope this helps!
Answer:
161 mL
Explanation:
- Pb(NO₃)₂(aq) + Ba(OH)₂(aq) → Pb(OH)₂(s) + Ba(NO₃)₂(aq)
First we <u>calculate how many Pb⁺² moles reacted</u>, using the<em> given concentration and volume of the Pb(NO₃)₂ solution</em>:
- 163 mL * 0.656 M = 107 mmol Pb(NO₃)₂
As<em> 1 millimol of Pb(NO₃)₂ would react with 1 millimol of Ba(OH)₂,</em> to precipitate 107 mmoles of Pb(NO₃)₂ we would require 107 mmoles of Ba(OH)₂.
Using the number of moles and the concentration we can <u>calculate the required number of milliliters</u>:
- 0.666 M = 107 mmol / x mL
To balance Ca + Cl2 = CaCl2 you'll need to be sure to count all of atoms on each side of the chemical equation.
Once you know how many of each type of atom you can only change the coefficients (the numbers in front of atoms or compounds) to balance the equation for Calcium + Chlorine gas.
Answer:
The Enthalpy of neutralization
Explanation:
The reaction of a base (KOH) with an acid (HCl) produce water and its salt (KCl) is called <em>Neutralization Reaction.</em>
This neutralization releases 57kJ/mol.
As the type of enthalpy is due the type of reaction. This enthalpy is:
<h3>The Enthalpy of neutralization</h3>
Given reaction:
N2(g) + 3H2(g) → 2NH3(g)
The standard free energy change is given as:
ΔG° = ∑ nΔGf(products) - ∑ nΔGf(reactants)
= [2ΔGf(NH3(g))] - [ΔGf(N2(g)) + 3ΔGf(H2(g))]
= [2(-16.48)] - [ 1(0) + 3(0)] = -32.96 kJ
Ans: Free energy of the reaction is -32.96 kJ, i.e. reaction is spontaneous.
