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Mrrafil [7]
3 years ago
5

The last element in any period always has:

Chemistry
2 answers:
sergeinik [125]3 years ago
8 0

Answer:

(A) eight electrons in the outermost energy level.

(C) the properties of a noble gas.

Explanation: option A and C are correct !

brilliants [131]3 years ago
4 0

Answer: The correct option is The properties of a noble gas.

Explanation: There are 7 periods in the periodic table.

The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).

  • The electronic configuration for Helium is 1s^2. For He, The outermost electrons are 2.
  • The electronic configuration for all the other elements is ns^2ns^6 ( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.

All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.

Therefore, the last element of each period always have the properties of a noble gas.

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A 40.2 g sample of a metal heated to 99.3°C is placed into a calorimeter containing 120 g of water at 21.8°C. The final temper
aliina [53]

Answer:

B) Iron (c=0.45 J/g°C)

Explanation:

Given that:-

Heat gain by water = Heat lost by metal

Thus,  

m_{water}\times C_{water}\times (T_f-T_i)=-m_{metal}\times C_{metal}\times (T_f-T_i)

Where, negative sign signifies heat loss

Or,  

m_{water}\times C_{water}\times (T_f-T_i)=m_{metal}\times C_{metal}\times (T_i-T_f)

For water:

Mass = 120 g

Initial temperature = 21.8 °C

Final temperature = 24.5 °C

Specific heat of water = 4.184 J/g°C

For metal:

Mass = 40.2 g

Initial temperature = 99.3 °C

Final temperature = 24.5 °C

Specific heat of metal = ?

So,  

120\times 4.184\times (24.5-21.8)=40.2\times C_{metal}\times (99.3-24.5)

40.2C_{metal}\left(99.3-24.5\right)=120\times \:2.7\times \:4.184

40.2C_{metal}\left(99.3-24.5\right)=1355.616

C_{metal}=0.45\ J/g^0C

<u>This value corresponds to iron. Thus answer is B.</u>

3 0
3 years ago
What is the major organic product obtained from the following reaction?
trapecia [35]

The correct option is (b)

NaNH2 is an effective base. It can be a good nucleophile in the few situations where its strong basicity does not have negative side effects. It is employed in elimination reactions as well as the deprotonation of weak acids.Alkynes, alcohols, and a variety of other functional groups with acidic protons, such as esters and ketones, will all be deprotonated by NaNH2, a powerful base.Alkynes are deprotonated with NaNH2 to produce what are known as "acetylide" ions. These ions are powerful nucleophiles that can react with alkyl halides to create carbon-carbon bonds and add to carbonyls in an addition reaction.Acid/base and nucleophilic substitution are the two types of reactions.Using the right base, terminal alkynes can be deprotonated to produce a carbanion.A good C is the acetylide carbanion.The acetylide carbanion can undergo nucleophilic substitution reactions because it is a potent C nucleophile. (often SN2) with 1 or 2 alkyl halides with electrophilic C to create an internal alkyne (Cl, Br, or I).Elimination is more likely to occur with 3-alkyl halides.It is possible to swap either one or both of the terminal H atoms in ethylene (acetylene) to create monosubstituted (R-C-C-H) and symmetrical (R = R') or unsymmetrical (R not equal to R') disubstituted alkynes (R-C-C-R').

Learn more about NANH2 here :-

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7 0
2 years ago
If it is 9 am in San Francisco, is it also 9 am in New York? Why or why not?
lys-0071 [83]

Answer:

san franciso is 3 hours beind and with NewYork

Explanation:

5 0
2 years ago
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What are the steps to go from the names of compounds to the formulas
marysya [2.9K]

Answer:

Explanation:

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4 0
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Genrish500 [490]

Answer:

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Explanation:

4 0
3 years ago
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