That stress is compression. It forms Reverse Faults.
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
Answer is: 162 g of beryllium.
Chemical reaction: Be + 2H₂O → Be(OH)₂ + H₂.
m(H₂) = 36,0 g.
m(Be) = ?
n(H₂) = m(H₂) ÷ M(H₂)
n(H₂) = 36,0 g ÷ 2 g/mol = 18 mol.
from reaction: n(Be) : n(H₂) = 1 : 1.
n(Be) = n(H₂) = 18 mol.
m(Be) = n(Be) · M(Be).
n(Be) = 18 mol · 9 g/mol = 162 g.
Answer: Elements in Group 2
Explanation: The periodic table was arranged by Dmitri Mendeleev specifically around similarites in their chemical behaviors. He found that as atomic number increases, at some point an element starts to react in a manner similar to a previous one. When that happened, he would place the larger element under the smaller one, and eventually noticed a periodicity in the table. Elements in a column (Groups) had similiar chemical properties. We know today that these similarities are due to the electron configuration, and that these configurations repeat themselves. He left gaps in the table when he could find an existing element with properties similar to others in that group. I big leap of faith, but it worked. Elements for those missing boxes were eventually discovered.
