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AysviL [449]
3 years ago
8

Why are covalently bonded materials generally less dense than metallically or ionically bonded ones

Chemistry
1 answer:
Snowcat [4.5K]3 years ago
3 0

Explanation:

It is known that ionic compounds are formed by transfer of electrons from one atom to another. And, during this type of bonding there occurs long range of bonding structure due to the nature of ionic bonds.

As these also contain opposite charge hence, they are more tightly packed with each other.

Whereas in covalent bonding the atoms tend to share electrons due to which they are not tightly packed with each other. As a result, empty spaces are created between the bonding structure which tends to reduce the density of the material.

Therefore, covalently bonded materials generally less dense than metallically or ionically bonded ones.

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Kepler did not study the speed of the planets, rather, he studied how the planets move in the solar system. He proposed three laws. As a summary, he described that the planets move around the sun in the shape of an ellipse (orbit), and the Sun being one of the foci. Then, he proposed the period for the planet to complete one revolution around the Sun. 

On the other hand, Newton studied the forces acting on the planet (or any object in space) that explain how the planets move around the solar system as described by Kepler. Also, Kepler's observations only apply to planets and not the moons or satellites. Thus, Kepler only made laws from observations, while Newton based it from underlying principles that led him to mathematical equations such as the law of universal gravitation.
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3 years ago
Read 2 more answers
When 0.620 gMngMn is combined with enough hydrochloric acid to make 100.0 mLmL of solution in a coffee-cup calorimeter, all of t
OleMash [197]

Answer:

The enthalpy change during the reaction is -199. kJ/mol.

Explanation:

Mn(s)+2HCl(aq)\rightarrow  MnCl_2(aq)+H_2(g)

Mass of solution = m

Volume of solution = 100.0 mL

Density of solution = d = 1.00 g/mL

m=1.00 g/mL\times 100.0 mL = 100 g

First we have to calculate the heat gained by the solution in coffee-cup calorimeter.

q=m\times c\times (T_{final}-T_{initial})

where,

m = mass of solution = 100 g

q = heat gained = ?

c = specific heat = 4.18 J/^oC

T_{final} = final temperature = 23.1^oC

T_{initial} = initial temperature = 28.9^oC

Now put all the given values in the above formula, we get:

q=100 g \times 4.18 J/^oC\times (28.9-23.1)^oC

q=2,242.4 J=2.242 kJ

Now we have to calculate the enthalpy change during the reaction.

\Delta H=-\frac{q}{n}

where,

\Delta H = enthalpy change = ?

q = heat gained = 2.242 kJ

n = number of moles fructose = \frac{\text{Mass of manganese}}{\text{Molar mass of manganese}}=\frac{0.620 g}{54.94 g/mol}=0.0113 mol

\Delta H=-\frac{2.242 kJ}{0.0113 mol }=-199. kJ/mol

Therefore, the enthalpy change during the reaction is -199. kJ/mol.

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Answer

they atract

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1 milliliter of water (ml)

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