Answer:
70mol
Explanation:
The equation of the reaction is given as:
2C₂H₂ + 5O₂ → 4CO₂ + 2H₂O
Given parameters:
Number of moles of acetylene = 35.0mol
Number of moles of oxygen in the tank = 84.0mol
Unknown:
Number of moles of CO₂ produced = 35.0mol
Solution:
From the information given about the reaction, we know that the reactant that limits this combustion process is acetylene. Oxygen is given in excess and we don't know the number of moles of this gas that was used up. We know for sure that all the moles of acetylene provided was used to furnish the burning procedure.
To determine the number of moles of CO₂ produced, we use the stoichiometric relationship between the known acetylene and the CO₂ produced from the balanced chemical equation:
From the equation:
2 moles of acetylene produced 4 moles of CO₂
∴ 35.0 mol of acetylene would produced:
= 70mol
<span>There are three atoms of Sn (Stannous or Tin) in</span> 356.13 g of Sn.
<span>One atom of Sn has the atomic mass (m</span>ₐ<span>) of </span>118,71u which means:
356.13/118.71=3 atoms of Sn
The mass number (symbol A) also called atomic mass number or nucleon number is the total number of protons and neutrons in an atomic nucleus. It determines the atomic mass of atoms and it is in the periodic table.
Answer:
1. 
2. 
Explanation:
An ion is formed when an atom that is said to be neutral gains or losses electrons.
It is thought that a negative ion (anion) is produced as it gains electrons and a positive ion (cation) is formed when it loses an electron.
Atomic number is the total number of protons and electrons in a neutral atom.
From the information
Protons = 14
electron = 18
Net Charge = no of proton - no of electron
= 14 - 18 = -4
Mass number = 14 + 15 = 29
Thus, the chemical symbol = 
For ion with 27 proton, 32 neutrons and 25 electrons
Net charge = 27 - 25 = +2
Mass number = 27 + 32 = 59
Thus, the chemical symbol = 
The anode is the electrode where the oxidation occurs.
Cathode is the electrode where the reducction occurs.
Equations:
Mn(2+) + 2e- ---> Mn(s) Eo = - 1.18 V
2Fe(3+) + 2e- ----> 2 Fe(2+) 2Eo = + 1.54 V
The electrons flow from the electrode with the lower Eo to the electrode with the higher Eo yielding to a positive voltage.
Eo = 1.54 V - (- 1.18) = 1.54 + 1.18 = 2.72
Answer: 2.72 V