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SVETLANKA909090 [29]
3 years ago
5

What type of map should have a north arrow, key ,and map scales?

Chemistry
1 answer:
KengaRu [80]3 years ago
7 0
Your answer would be C


hope this helped
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Described the role of electrons in the formation of a covalent bond​
My name is Ann [436]

Sharing of two electrons make a <u>Covalent </u>bond.

<u>Explanation: </u>

Attractions among the atoms bring them together. So the electrons from each of the atoms are attracted towards the nucleus of those two atoms, that “share” the electrons produces a covalent bond.

It is also named as molecular bond,  a bond that entails the sharing of a pair of electrons among the atoms. When the atoms share the electrons among themselves, it produces a molecule, which is more stable than the atom.

If the attractions between the atoms are strong enough and if every atom has enough space for the electrons in its outermost energy level then there occurs covalent bonding. So electrons are very important in the covalent bond formation.

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3 years ago
If an endothermic reaction occurs in solution, the temperature of the solution will
Ahat [919]

Answer:

increase

Explanation:

because in endothermic reaction heat is absorbed

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3 years ago
What do the elements in each pair have in common? k, kr be, mg ni, tc b, ge al, pb?
tensa zangetsu [6.8K]
The pairs are:
K, Kr - Same period
Be, Mg - Same group
Ni, Tc - Both are transition metals
B, Ge - Both are metaloids
Al, Pb - Both form inert oxides
4 0
3 years ago
HELP ME PLS I NEED HELP I HAVE 5MINS
Leviafan [203]

Answer:

K

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2 years ago
For a process Arightwards harpoon over leftwards harpoonB, at 25 °C there is 10% of A at equilibrium while at 75 °C, there is 80
Lostsunrise [7]

This question is describing the following chemical reaction at equilibrium:

A\rightleftharpoons B

And provides the relative amounts of both A and B at 25 °C and 75 °C, this means the equilibrium expressions and equilibrium constants can be written as:

K_1=\frac{90\%}{10\%}=9\\\\K_2=\frac{20\%}{80\%}  =0.25

Thus, by recalling the Van't Hoff's equation, we can write:

ln(K_2/K_1)=-\frac{\Delta H}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Hence, we solve for the enthalpy change as follows:

\Delta H=\frac{-R*ln(K_2/K_1)}{(\frac{1}{T_2} -\frac{1}{T_1} ) }

Finally, we plug in the numbers to obtain:

\Delta H=\frac{-8.314\frac{J}{mol*K} *ln(0.25/9)}{[\frac{1}{(75+273.15)K} -\frac{1}{(25+273.15)K} ] } \\\\\\\Delta H=4,785.1\frac{J}{mol}

Learn more:

  • brainly.com/question/10038290
  • brainly.com/question/19671384
5 0
2 years ago
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