Answer:
Kc = 12.58
Explanation:
Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3
Kc = (0.052441)(0.10513)/(0.002385)(0.18373)
Kc = 0.0005513/0.000438
Kc = 12.58
Hope that helps!!
Answer:
B. CH3CH2OH
Explanation:
Ethanol has a chemical formula of CH3CH2OH, it is the second member of the series in the alkanol family. Ethanol is a colourless, volatile liquid with a characteristic smell and taste. It is readily soluble in water in all proportions. It has a boiling point of 78° C. The physical properties such as the solubility of alkanols are affected by the presence of hydrogen bonding. The hydroxyl group is capable of bonding to other alkanol molecules. The boiling points rise with increasing molecular mass.
Hydrogen bonding helps the molecules to stick together. For example comparing the boiling point of pentane ( 36° C) with that of butan-1-ol (118° C) , the boiling point of alkanol is much higher even though the two compound are of similar relative molecular mass. This is due to the presence of hydrogen bonds in butanol.
Hydrocarbons are not soluble in water but alkanols are soluble in water because of the hydroxyl groups in the molecules can form hydrogen bond with water. Solubility of alkanol in water decreases as the number of carbon atom increases. Primary alcohol with more than five carbon atoms are insoluble in water.
Answer : The steps for balancing a redox reaction using half reaction method will include the following steps :
1) Write the basic ionic form of the reaction equation.
2) Divide the equation into two separate half reaction which will have oxidation half and reduction half reactions.
3) Balance the atoms present in each half of the reactions except O and H atoms in the reaction.
4) Balance the oxygen atoms in the reduction half of the reaction by the addition of two water molecules in the given conditions of the reaction.
5) Balance the charges present in both the half reactions.
6) The two halves of the equations reduction and oxidation are added to complete the overall reactions. After the addition of two reaction halves, cancel the same number electrons on both sides. The net ionic equation can be written after this step.
7) Finally, Verify whether the equation consists of the same type, number of co-efficients, and charges on both sides of the equation.
The volume of HCOOH is 0.003 Lt and the volume of HCOONa is 0.017 LT.
Explanation:
For acidic buffer
PH = Pka + log
4.25 = 3.75 + log
log = 0.75
log = 5.62 is considered as equation 1.
Let V₁ ml of HCOOH and V₂ml of HCOONA was mixed and the final volume is 100 ml .
V₁ *0.5 + V₂ *0.5 =0.1 * 100
V₁+V₂ = = 20 is the second equation.
Now concentration of salt and acid are
V₂ [salt] = 0.5/100 and v₁ [acid] = 0.5/100
by substituting these in equation 1 we get,
= 5.62
V₂ = V₁ * 5.62 is the 3 rd equation
by solving 2 and 3 we get,
5.62 V₁ + V₁ = 20
= 6.62 V₁ = 20
V₁ = 3.02 ml
V₁ = 0.003 LT
V₂ = V₁ * 5.62
V₂ = 0.017 LT
Hence the volume of HCOOH is 0.003 Lt and the volume of HCOONa is 0.017 LT.
Answer:
H-CL
Explanation:
Hydrochloric acidis an acid wiith a diatomic molecule, consisting of a hydrogen atom H and a chlorine atom Cl connected by a polar covalent bond