A.)Neutral. Since amount of protons and electron are equal and amount of protons cannot be changed so it is stable or neutral.
Answer:
pH = 4.17
Explanation:
According to the molar concentration you stated, pH of the solution is: 4.17
Remember that pH = - log [H⁺]
and [H⁺] = 10^-pH
When:
pH > 7 → Basic solution
pH = 7 → Neutral solution
pH < 7 → Acid solution
Answer:
See explanation
Explanation:
The molecular geometry of an atom is connected to the number of electron pairs that surround it(whether lone pairs or bonding pairs) as well as its hybridization state. We shall now examine the N, P, or S atoms in each of the following compounds.
a)
In H3PO4, P has a tetrahedral molecular geometry and is sp3 hybridized.
b) In NH4NO3
N is sp3 hybridized in NH4^+ and sp2 hybridized in NO3^-. Also, N is tetrahedral in NH4^+ but trigonal planar in NO3^-.
c) In S2Cl2, we expect a tetrahedral geometry but as a result of the presence of two lone pairs on each sulphur atom, the molecular geometry is bent. The sulphur is sp3 hybridized.
d) In K4[O3POPO3], each phosphorus atom is in a tetrahedral molecular geometry and is sp3 hybridized.
Answer:
Only Reaction 1
Explanation:
In reaction 1, there is a change in state from solid to liquid. Hence, there is an increase in number of ways particles and their energies could be arranged. As a result, entropy increases.
In reaction 2, there is a decrease in amount of gas particles (4 mol to 2 mol). Hence there is a decrease in the number of ways particles and their energies could be arranged. As a result entropy decreases
The question is incomplete, the complete question is;
1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?
A. CL2
B. SF6
C. Kr
D. UF6
E. Xe
Answer:
SF6
Explanation:
From Graham's law;
Let the rate of diffusion of oxygen be R1
Let the rate of diffusion of unknown base be R2
Let the molar mass of oxygen by M1
Let the molar mass of unknown gas be M2
Hence;
R1/R2 = √M2/M1
So;
2.14/1 = √M2/32
(2.14/1)^2 = M/32
M= (2.14/1)^2 × 32
M= 146.6
This is the molar mass of SF6 hence the answer above.
