In 1871, a Russian Chemist, Dimitri Mendeleev, gave a useful scheme for classification of elements. He presented the first regular periodic table in which elements of similar chemical properties were arranged in eight vertical columns called groups. The horizontal rows of table were called periods. He arranged elements in ascending order of their atomic masses and found that elements having similar chemical properties appeared at regular intervals. This observation was called Periodic Law.
15.63 mol. You need 15.63 mol HgO to produce 250.0 g O_2.
<em>Step 1</em>. Convert <em>grams of O_2 to moles of O_2</em>
Moles of O_2 = 250.0 g O_2 × (1 mol O_2/32.00 g O_2) = 7.8125 mol O_2
<em>Step 2</em>. Use the molar ratio of HgO:O_2 to convert <em>moles of O_2 to moles of HgO
</em>
Moles of HgO = 0.8885 mol O_2 × (2 mol HgO/1 mol O_2) = <em>15.63 mol HgO</em>
<h3>
Answer:</h3>
20.62 Kilo-joules
<h3>
Explanation:</h3>
- The Enthalpy of combustion of ethyl alcohol is -950 kJ/mol.
- This means that 1 mole of ethyl alcohol evolves a quantity of heat of 950 Joules when burned.
Molar mass of ethyl ethanol = 46.08 g/mol
Therefore;
46.08 g of C₂H₅OH evolves heat equivalent to 950 kilojoules
We can calculate the amount of heat evolved by 1 g of C₂H₅OH
Heat evolved by 1 g of C₂H₅OH = Molar enthalpy of combustion ÷ Molar mass
= 950 kJ/mol ÷ 46.08 g/mol
= 20.62 Kj/g
Therefore, a gram of C₂H₅OH will evolve 20.62 kilo-joules of heat
They do not show the same season. one is faced a different part of the sun
Answer:
_5_ AsO2−(aq) + 3 Mn2+(aq) + _2_ H2O(l) → _5_ As(s) + _3_ MnO4−(aq) + _4_ H+(aq)
Explanation:
Step 1:
The unbalanced equation:
AsO2−(aq) + 3 Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
Step 2:
Balancing the equation.
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + MnO4−(aq) + H+(aq)
The above equation can be balanced as follow:
There are 3 atoms of Mn on the left side of the equation and 1 atom on the right side. It can be balance by putting 3 in front of MnO4− as shown below:
AsO2−(aq) + 3Mn2+(aq) + H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 12 atoms of O on the right side and a total of 3 atoms on the left side. It can be balance by putting 5 in front of AsO2− and 2 in front of H2O as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + H+(aq)
There are 4 atoms of H on the left side and 1 atom on the right side. It can be balance by putting 4 in front of H+ as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → As(s) + 3MnO4−(aq) + 4H+(aq)
There are 5 atoms of As on the left side and 1 atom on the right side. It can be balance by putting 5 in front of As as shown below:
5AsO2−(aq) + 3Mn2+(aq) + 2H2O(l) → 5As(s) + 3MnO4−(aq) + 4H+(aq)
Now the equation is balanced
