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ELEN [110]
3 years ago
8

Laws differ from theories because laws do not provide

Chemistry
2 answers:
natita [175]3 years ago
4 0
The answer is C! Trust Me!
Angelina_Jolie [31]3 years ago
3 0
C. Explanation since laws are already proven and doent need to be proven further
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For each of the following sets of elements, circle the element expected to be most electronegative and draw a box which is expec
Evgen [1.6K]

Answer: The element expected to be most electronegative is Ca.

The element expected to be least electronegative is K.

Explanation:

Electronegativity is defined as the property of an element to attract a shared pair of electron towards itself.

Down the group:

The size of an atom increases as we move down the group because a new shell is added and electron gets added up.

As, the size of an element increases, the valence electrons gets away from the nucleus. So, the attraction between the nucleus and the shared pair of electrons decreases

Hence, electronegativity decreases moving from top to bottom down a group

Across a period:

The size of an atom decreases as we move across the period because the electrons get added to the same shell and the nuclear charge keeps on increasing. Thus the electrons get more tightly held by the nucleus.

As, the size of an element decreases, the valence electrons come near to the nucleus. So, the attraction between the nucleus and the shared pair of electrons increases.

Hence, electronegativity increases moving across left to right in a period.

Thus as K, Sc and Ca are arranged across a period, the electronegativity order is K< Sc < Ca.

7 0
3 years ago
Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)
Dovator [93]

Answer:

1. 0.97 V

2. Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)

Explanation:

In this case, we can start with the <u>half-reactions</u>:

Ag^+~_(_a_q_)->~Ag_(_s_)

Al_(_s_)~->~Al^+^3~_(_a_q_)

With this in mind we can <u>add the electrons</u>:

Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)  <u>Reduction</u>

Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~ <u>Oxidation</u>

The reduction potential values for each half-reaction are:

Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_) - 0.69 V

Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_) -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to <u>flip</u> the reduction potential value:

Al_(_s_)~->~Al^+^3~+~2e^-~ +1.66 V

Finally, to calculate the overall potential we have to <u>add</u> the two values:

1.66 V - 0.69 V = <u>0.97 V</u>

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)

I hope it helps!

3 0
3 years ago
The half-life of cobalt-60 is 5.3 years. after __________ years, 1/4 of the original amount of cobalt-60 will remain.
cestrela7 [59]
The  number of  years  required  for 1/4  cobalt-60  to remain   after   decay  is calculated  as follows

  after  one  half life  1/2 of the original  mass isotope  remains

after  another half life 1/4  mass  of  original  mass  remains

therefore  if one  half  life is  5.3   years  then  the  years required

= 2  x 5.3years =  10.6   years
4 0
2 years ago
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A 5.0 g piece of metal at 100c is cooled to 15c.It releases 318.3 J of heat in the process.Calculate the specific heat of the me
Mnenie [13.5K]

Answer:

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5 0
2 years ago
Two moles of magnesium and five moles of oxygen are placed in a vessel. When magnesium is ignited ca two moles of magnesium and
Greeley [361]
Limiting reactant in this experiment would be Magnesium since it will run out first
4 0
3 years ago
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