Explanation:
Charges on both magnesium and oxygen is 2. Though opposite in sign, they have equal charges so, both of them will be cancelled by each other.
As a result, formula of magnesium oxide is MgO and not 
.
The student write the equation as 
, it is not correct.
Therefore, given equation will be balanced as follows.
Since, number of atoms on both reactant and product side are equal. Hence, this equation is completely balanced.
Answer:
So sorry if I was wrong but I think it's B. Because from the source states,
https://socratic.org/questions/which-group-on-the-periodic-table-is-the-least-reactive-why
"The least reactive elements are those who have a full outermost valence shell ie they have 8 electrons in the outer shell so elements such as helium, neon, radon or the transition elements."
Answer:
I would expect the gas rate determined in this manner to be too low
Explanation:
A Rotameter can be designed to respond to the sensitivity of density, velocity, to measure the flow rate of liquid or gas enclosed in a tube. Liquids are denser than gas, and since the gas rate to be determined needed to respond to the velocity head alone of the rotameter so as to bring the forces in the tube equilibrium. Knowing if there is no flow, then the float would remain at the bottom, so gas has to flow at a higher rate compared to the liquid so the float would be in a similar position making it easier to measure the flowrate. This leaves the gas rate to be determined too low.
Answers and Explanation:
a)- The chemical equation for the corresponden equilibrium of Ka1 is:
2. HNO2(aq)⇌H+(aq)+NO−2
Because Ka1 correspond to a dissociation equilibrium. Nitrous acid (HNO₂) losses a proton (H⁺) and gives the monovalent anion NO₂⁻.
b)- The relation between Ka and the free energy change (ΔG) is given by the following equation:
ΔG= ΔGº + RT ln Q
Where T is the temperature (T= 25ºc= 298 K) and R is the gases constant (8.314 J/K.mol)
At the equilibrium: ΔG=0 and Q= Ka. So, we can calculate ΔGº by introducing the value of Ka:
⇒ 0 = ΔGº + RT ln Ka
ΔGº= - RT ln Ka
ΔGº= -8.314 J/K.mol x 298 K x ln (4.5 10⁻⁴)
ΔGº= 19092.8 J/mol
c)- According to the previous demonstation, at equilibrium ΔG= 0.
d)- In a non-equilibrium condition, we have Q which is calculated with the concentrations of products and reactions in a non equilibrium state:
ΔG= ΔGº + RT ln Q
Q= ((H⁺) (NO₂⁻))/(HNO₂)
Q= ( (5.9 10⁻² M) x (6.7 10⁻⁴ M) ) / (0.21 M)
Q= 1.88 10⁻⁴
We know that ΔGº= 19092.8 J/mol, so:
ΔG= ΔGº + RT ln Q
ΔG= 19092.8 J/mol + (8.314 J/K.mol x 298 K x ln (1.88 10⁻⁴)
ΔG= -2162.4 J/mol
Notice that ΔG<0, so the process is spontaneous in that direction.
