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Alex777 [14]
3 years ago
8

In which group of the Periodic Table do most of the elements exhibit both positive and negative oxidation states?

Chemistry
1 answer:
Shtirlitz [24]3 years ago
4 0
C i think but you should pick it anyway
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Please help if you can !
iogann1982 [59]

Answer:

7630

Explanation:

8 0
3 years ago
Read 2 more answers
Identify the hybridization of each carbon atom for the molecule above
ipn [44]

Carbons starting from the left end:

  1. sp²
  2. sp²
  3. sp²
  4. sp
  5. sp

Refer to the sketch attached.

<h3>Explanation</h3>

The hybridization of a carbon atom depends on the number of electron domains that it has.

Each chemical bond counts as one single electron domain. This is the case for all chemical bonds: single, double, or triple. Each lone pair also counts as one electron domain. However, lone pairs are seldom seen on carbon atoms.

Each carbon atom has four valence electrons. It can form up to four chemical bonds. As a result, a carbon atom can have up to four electron domains. It has a minimum of two electron domains, with either two double bonds or one single bond and one triple bond.

  • A carbon atom with four electron domains is sp³ hybridized;
  • A carbon atom with three electron domains is sp² hybridized;
  • A carbon atom with two electron domains is sp hybridized.

Starting from the left end (H₂C=CH-) of the molecule:

  • The first carbon has three electron domains: two C-H single bonds and one C=C double bond; It is sp² hybridized.
  • The second carbon has three electron domains: one C-H single bond, one C-C single bond, and one C=C double bond; it is sp² hybridized.
  • The third carbon has three electron domains: two C-C single bonds and one C=O double bond; it is sp² hybridized.
  • The fourth carbon has two electron domains: one C-C single bond and one C≡C triple bond; it is sp hybridized.
  • The fifth carbon has two electron domains: one C-H single bond and one C≡C triple bond; it is sp hybridized.

8 0
3 years ago
A metal pellet with a mass of 100.0 g, originally at 116°C, is dropped into a cup of water, initially at
Alina [70]

Answer:

C, 42g

Explanation:

In thermal equilibrium, both bodies (metal pellet and water) both have the same final temperature (46.3°C).

Assuming no heat is lost to surroundings,

the energy lost from metal pellet = energy gained for water

Since E = mc∆T

(energy = mass x specific heat capacity x temperature change)

mc∆T (metal pellet) = mc∆T (water)

100 x 0.568 x (116-46.3) = m 4.184 (46.3 - 23.8)

3958.96 = 94.14m

m = 42g

6 0
2 years ago
What is the density of an 84.7 g sample of an unknown substance if the sample occupies 49.6cm cubed
Ahat [919]

Answer:

             1.70 g.cm⁻³

Solution:

Data Given;

                   Mass  =  84.7 g

                   Volume  =  49.6 cm³

                   Density  =  ?

Formula Used;

                   Density  =  Mass ÷ Volume

Putting values,

                   Density  =  84.7 g ÷ 49.6 cm³ 

                   Density  =  1.70 g.cm⁻³

7 0
3 years ago
Hiii pls help me to write out the ionic equation ​
emmasim [6.3K]

Answer:

<u>STEP I</u>

This is the balanced equation for the given reaction:-

2KOH_{(aq)} + H_2SO_4{}_{(aq)}   \rightarrow K_2SO_4{}_{(aq)} + 2H_2O_{(l)}

<u>STEP II</u>

The compounds marked with (aq) are soluble ionic compounds. They must be

broken into their respective ions.

see, in the equation KOH, H2SO4, and K2SO4 are marked with (aq).

On breaking them into their respective ions :-

  • 2KOH -> 2K+ + 2OH-
  • H2SO4 -> 2H+ + (SO4)2-
  • K2SO4 -> 2K+ + (SO4)2-

<u>STEP III</u>

Rewriting these in the form of equation

\underline{\pmb{2K^+} }+ 2OH^- + 2H^+ + \pmb{\underline{{SO_4{}^{2-}}} \: \rightarrow \:  \underline{\pmb{2K^+}}} + \underline{\pmb{SO_4{}^{2-}}} + 2H_2O

<u>STEP </u><u>IV</u>

Canceling spectator ions, the ions that appear the same on either side of the equation

<em>(note: in the above step the ions in bold have gotten canceled.)</em>

\boxed{ \mathfrak{ \red{ 2OH^-{}_{(aq)} + 2H^+{(aq.)} \rightarrow H_2O{}_{(l)}}}}

This is the net ionic equation.

____________________________

\\

\mathfrak{\underline{\green{ Why\: KOH \:has\:  been\: taken\: as\: aqueous ?}}}

  • KOH has been taken as aqueous because the question informs us that we have a solution of KOH. by solution it means that KOH has been dissolved in water before use.

[Alkali metal hydroxides are the only halides soluble in water ]

4 0
3 years ago
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