Answer:
(1) The two column proof is presented here as follows;
Statement
Reasons
1. C is the midpoint of
Given
B is the midpoint of
D is the midpoint of
2. AC = CE, AB = BC, CD = DE
Definition of midpoint
3. AB + BC = AC, CD + DE = CE
Segment addition postulate
4. CD + DE = AC
Substitution property of equality
5. AB + BC = CD + DE
Substitution property of equality
6. BC + BC = CD + CD
Substitution property of equality
7. 2·BC = 2·CD
Addition of two identical quantities
8. BC = CD
Division property of equality
9. BC ≅ CD
Definition of Congruence
(2) The two column proof for the triangular geometric figure is presented here as follows;
Statement
Reasons
1. 2·WV = XY; 2·YZ = XW; WV = YZ
Given
2. 2·WV = 2·YZ
Multiplication property
3. XY = XW
Substitution Property
4. XW + WV = XV
Segment Addition Property
XY + YZ = XZ
5. XW + WV = XV
Substitution Property
XW + WV = XZ
6. XV = XZ
Substitution Property
Explanation:
Answer:
iodine
Explanation:
<em>The substance that can break easily from the list of substances is the </em><em>iodine</em><em>.</em>
<u>Iodine exists as a molecular solid and hence, the molecules are held together by weak Van Der Waal's forces. Iodine thus has characteristic properties of typical molecular solids which include being able to be broken into pieces easily (brittle), low melting and boiling points, insolubility in water, among other properties.</u>
The characteristics of iodine are unlike sodium, calcium, and copper that exist as metal which is hard, tough with high tensile strengths.
Answer:
PI = 6.04
In the attached figure are the missing hydrogen atoms
Explanation:
Given both pk, the isoelectric point will be equal to:
PI = (1/2)*(pka + pka2), where pka = 2.32 and pka2 = 9.76
PI = (1/2)*(2.32 + 9.76) = 6.04
Answer:
Reaction 1 - elimination
Reaction 2- substitution
Reaction 3- the two combined
Explanation:
In the first reaction, a strong base, KOH is used in the reaction. Remember that when a tertiary alkyl halide is the substrate, the use of a strong base leads to the domination of E1 mechanism over SN1 mechanism, hence the product shown in the image attached.
In reaction 2, the secondary alkyl halide must undergo an SN2 substitution since an aprotic solvent (THF) is used. This leads to inversion of configuration as shown in the product.
In reaction 3, CN^- is both a strong base and a good nucleophile hence a mixture of substitution and elimination products are formed depending on the nature of the alkyl halide.