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algol13
3 years ago
10

Name two objects which are hard to touch​

Chemistry
2 answers:
ziro4ka [17]3 years ago
7 0

fire and mercury

i think i am not sure

but pls brainlist

Anni [7]3 years ago
3 0

Answer:

Name of two hard objects is Brick and Wood.

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A neutral atom with the electron configuration 2-8-6 would most likely form a bond with an atom having the configuration
Fittoniya [83]

Answer:

The configuration of the atom would be 2-8-2.

Explanation:

Any atom of an element combines with other element to complete its octet and become stable.

The electron configuration of the given atom is 2-8-6. That means the atom has 6 electrons in its outermost shell. To become stable the atom should have 8 electrons in its outermost shell. The given atom has 6 electrons so it either lose 6 electrons or gain 2 electrons to complete its octet.

But we know the atom having 5,6,7 electrons in its outermost shell they do not lose, they gain either 3 or 2 or 1 electrons to complete its octet.

So we say that atom with the electron configuration 2-8-6 bond with the atom having electron configuration 2-8-2.

8 0
3 years ago
The number of protons plus the number of neutrons equal the ______
NARA [144]
the answer is: Atomic number
7 0
2 years ago
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Bluebirds and blue jays, two bird species live in a small wooded area. The blue jays are very aggressive and eventually take ove
Dahasolnce [82]

Answer:

yes

Explanation:

it affects the bluebirds habitat

5 0
3 years ago
A change in the force of gravity on an object will affect its what
Gnoma [55]

A change in the force of gravity on an object will affect its weight.

4 0
2 years ago
When 0.0901 mol of an unknown hydrocarbon is burned in a bomb calorimeter, the calorimeter increases in temperature by 2.19°C. I
Alexus [3.1K]

Answer:

The heat of combustion for the unknown hydrocarbon is -29.87 kJ/mol

Explanation:

Heat capacity of the bomb calorimeter = C = 1.229 kJ/°C

Change in temperature of the bomb calorimeter = ΔT = 2.19°C

Heat absorbed by bomb calorimeter = Q

Q=C\times \Delta T

Q=1.229 kJ/^oC\times 2.19^oC=2,692 kJ

Moles of hydrocarbon burned in calorimeter = 0.0901 mol

Heat released on combustion = Q' = -Q = -2,692 kJ

The heat of combustion for the unknown hydrocarbon :

\frac{Q'}{0.090 mol}=\frac{-2,692 kJ}{0.0901 mol}=-29.87 kJ/mol

6 0
3 years ago
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