Answer:
The ground state electron configuration of ground state gaseous neutral tellurium is [Kr]. 4d10.
Answer:
184.62 ml
Explanation:
Let 
and 
be the initial and 
and 
be the final pressure, volume, and temperature of the gas respectively.
Given that the pressure remains constant, so
...(i)
= 200 ml
K
K
From the ideal gas equation, pv=mRT
Where p is the pressure, v is the volume, T is the temperature in Kelvin, m is the mass of air in kg, R is the specific gas constant.
For the initial condition,
For the final condition,
Equating equation (i), and (ii)
[from equation (i)]
Putting all the given values, we have
Hence, the volume of the gas at 3 degrees Celsius is 184.62 ml.
Answer:
2%
Explanation:
.98 is 98% of one and therefore they are missing 2%
Answer:
2HgS + 3O2 → 2HgO + 2SO2
The coefficients are: 2, 3, 2, 2
Explanation:
HgS + O2 → HgO + SO2
The equation can be balance as follow:
Put 3 in front of O2 as shown below:
HgS + 3O2 → HgO + SO2
Now we can see that there are 6 atoms of O on the left side of the equation and a total of 3 atoms on the right side. It can be balance by putting 2 in front of HgO and SO2 as shown below:
HgS + 3O2 → 2HgO + 2SO2
Now we have 2 atoms of both Hg and S on the right side and 1atom each on the left. It can be balance by putting 2 in front of HgS as shown below:
2HgS + 3O2 → 2HgO + 2SO2
Now the equation is balanced.
The coefficients are: 2, 3, 2, 2
The law of conservation of mass(matter) states that matter(mass) can neither be created nor destroyed during a chemical reaction but changes from one form to another. An unbalanced equation suggests that matter has been created or destroyed. While a balanced equation proofs that matter can never be created but changes to different form. This is the more reason we have count the atoms of an element on both side of the equation to see if they are balanced irrespective of the new form they assume in the product
Answer:
Explanation:
A sigma bond is formed when a hybrid orbital sp overlaps with another hybrid orbital or with s- or p- orbital
Ethylene has a structural formula of H - C≡ C- H
At the ground state; we have :
C | ⇅ | |↑ | ↑
2s 2p
At the excited state, we have:
C | ↑ | |↑ | ↑ | ↑
the hybrid orbital
C = | ↑ | ↑ |
2sp
