Answer:
1. HBr>HCl> H2S >BH3
2.K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
Explanation:
As one goes down a row in the Periodic Table the properties that determine the acid strength can be observed.
The atoms get larger in radius meaning that in strength, the strength of the bonds get weaker, conversely meaning that the acids get stronger.
For the halogen-containing acids above following the rows and periods, HBr has the strongest bond and is the strongest acid and others follow in this order.
HBr>HCl> H2S >BH3
Acid Dissociation Constant provides us with information known as the ionization constant which comes in handy to measure the acid's strength. The meaning of the proportions are thus, the higher the Ka value, the stronger the acid i.e. it liberates more number of hydrogen ions per mole of acid in solution.
In solution strong acids completely dissociate hence, the value of dissociation constant of strong acids is very high.
Following the cues above on Ka;
K_a1 very large — H2SO4
K_a1= 1.7 x 10^−2 — H2SO3
K_a1 = 1.7 x 10^−7 — H2S
For the conversions
I will start with pressure
1atm=101.3kPa
x =700kPa
x=700kPa/101.3kPa
x=6.91atm
Temperature
273K+30.00C
303K
Volume
1L=1000ml
x =50ml
x=0.05L
PV=nRT
6.91*0.05=n*0.08206*303
0.3455=24.86418n
0.3455/24.86418=n
0.0138=n
number of moles = 0.0138moles
Note: 0.08206 is the gas constant in this case
Answer:
Externalities are costs (negative externalities) or benefits (positive externalities), which are not reflected in free market prices. ... Market failure is a situation in which the free market leads to a misallocation of society's scarce ... and the killing of fish is not a cost that it would directly have to bear
Answer:
Frost Wedging - Hot and dry
Clay Formation - Cold and Wet
Dissolving - Cold, and dry
Explanation:
The frost wedging happen when the climatic condition is hot and dry. The dry weather compensates the heat and the rock wedging happens quickly. For clay formation the weather has to be cold and wet. The cold weather will make the sand indulge with rock particles resulting in the clay formation.
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol
