All categories

3 years ago

What type of solid has the highest melting point?

Chemistry

2 answers:

dem82 [27]3 years ago

8 0

Answer: B -Network solids

Ionic solids are held by positive and negative charged ions bonded by electrostatic forces. The electrostatic force is much stronger than dipole–dipole interactions, London dispersion forces, hydrogen bonding.

Molecular solids are held by dipole–dipole interactions, London dispersion forces, or hydrogen bonds. Benzene is an example of this. These inter-molecular force are much weaker than electrostatic force.

The metallic bonds are much weaker than electrostatic force. Similarly, in non-metallic solids the atoms are held by covalent bonds. These covalent bonds are weaker than the electrostatic force.

Thus we can conclude that electrostatic force is the strongest when compared to dipole–dipole interactions, London dispersion forces, hydrogen bonding,covalent and metallic bonds. Thus ionic solids will have the highest melting point as more energy is required to break the ionic bonds as this is the strongest bond compared to the other bonds.

erastova [34]3 years ago

3 0

Answer:

The correct answer option is B. network solid.

Explanation:

A network solid has the highest melting point.

The reason being that all of the atoms in this type of solid are covalently bonded to one another so the covalent bonds must be broken throughout the substance.

These covalent bonds holding the solid together are stronger than the ionic bonds. So it means that more heat energy is needed to break these bonds and therefore a higher temperature is required for it.

You might be interested in

So why di fishes not have legs\

marin [14]

Answer:

Fish need to be aerodynamic to move quickly. If they had legs they woulf be less aerodynamic and unable to get away from predators.

5 0

2 years ago

An electrolyte can be checked to ensure that it has correct chemical composition by measuring its _____.

UNO [17]

Answer:

specific gravity

Explanation:

3 0

2 years ago

Explain why the gamma rays do not bend.

Natalka [10]

They are electrically neutral, they are not deflected by magnetic fields

6 0

2 years ago

A 100 gram glass container contains 200 grams of water and 50.0 grams of ice all at 0°c. a 200 gram piece of lead at 100°c is ad

ASHA 777 [7]

$0 \; \textdegree{\text{C}}$

Explanation:

Assuming that the final (equilibrium) temperature of the system is above the melting point of ice, such that all ice in the container melts in this process thus

$E(\text{fusion}) = m(\text{ice}) \cdot L_{f}(\text{water}) = 66.74 \; \text{kJ}$ and
$m(\text{water, final}) = m(\text{water, initial}) + m(\text{ice, initial}) = 0.250 \; \text{kg}$

Let the final temperature of the system be $t \; \textdegree{\text{C}}$ . Thus $\Delta T (\text{water}) = \Delta T (\text{beaker}) = t(\text{initial}) - t_{0} = t \; \textdegree{\text{C}}$

$Q(\text{water}) &= &c(\text{water}) \cdot m(\text{water, final}) \cdot \Delta T (\text{water})= 1.047 \cdot t\; \text{kJ}$ (converted to kilojoules)
$Q(\text{container}) &= &c(\text{glass}) \cdot m(\text{container}) \cdot \Delta T (\text{container})= 0.0837 \cdot t \; \text{kJ}$
$Q(\text{lead}) &= &c(\text{lead}) \cdot m(\text{lead}) \cdot \Delta T (\text{lead})= 0.0255 \cdot (100 - t)\; \text{kJ}$

The fact that energy within this system (assuming proper insulation) conserves allows for the construction of an equation about variable $t$ .

$E(\text{absorbed} ) = E(\text{released})$

$E(\text{absorbed} ) = E(\text{fushion}) + Q(\text{water}) + Q(\text{container})$
$E(\text{released}) = Q(\text{lead})$

Confirm the uniformity of units, equate the two expressions and solve for $t$ :

$66.74 + 1.047 \cdot t + 0.0837 \cdot t = 0.0255 \cdot (80 - t)$

$t \approx -55.95\; \textdegree{\text{C}} < 0\; \textdegree{\text{C}}$ which goes against the initial assumption. Implying that the final temperature does not go above the melting point of water- i.e., $t \le 0 \; \textdegree{\text{C}}$ . However, there's no way for the temperature of the system to go below $0 \; \textdegree{\text{C}}$ ; doing so would require the removal of heat from the system which isn't possible under the given circumstance; the ice-water mixture experiences an addition of heat as the hot block of lead was added to the system.

The temperature of the system therefore remains at $0 \; \textdegree{\text{C}}$ ; the only macroscopic change in this process is expected to be observed as a slight variation in the ratio between the mass of liquid water and that of the ice in this system.

3 0

3 years ago

Please help me please please :(

Annette [7]

Answer:

B. and A.

Explanation:

May I have the Brainlliest award?

3 0

3 years ago