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3 years ago

How does a network solid differ from most other covalent compounds

Chemistry

1 answer:

blondinia [14]3 years ago

5 0

Answer:

Covalent compounds are held by intermolecular forces while network solids are held by strong bonds in unit cells which are closely packed together.

Explanation:

Covalent compound molecules are held by vanderwaals forces which are relatively weak but strong enough to hold some covalent molecules together in the solid state. However, network solids contain atom to atom covalent bonds arranged in an orderly manner and regular repeating unit cells to form a rigid three dimensional network solid.

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Explain how the mid-ocean ridge supports<br> the theory of sea floor spreading.<br> help plzzz:(

Sedaia [141]

The ridge of mid ocean b is the answer hope this helps

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4 years ago

Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr

RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$

The net reaction is :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant $K$ in terms of the equilibrium constants $K_1$ and $K_2$ . If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$ ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \ K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$ ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$ .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

4 0

3 years ago

Food dye allergies have increased in recent years. One study calculated that each 47.9 g serving of M&M's candy contains 240

77julia77 [94]

Answer:

Mass percent of food dyes = 0.0616%

Explanation:

Given data:

Mass of candy = 47.9 g

Calories = 240

Mass of fat = 10 g

Mass of carbohydrate = 34 g

Mass of protein = 2 g

Mass of food dyes = 29.5 mg

Mass percent of food dyes = ?

Solution:

First of all we will convert the mg into g.

Mass of food dyes = 29.5 mg × 1g /1000 mg = 0.0295 g

Mass percent of food dyes = mass of food dyes / total mass× 100

Now we will put the values.

Mass percent of food dyes = 0.0295 g / 47.9 g × 100

Mass percent of food dyes = 0.000616 × 100

Mass percent of food dyes = 0.0616%

3 0

3 years ago

Identify the false statement from the following.

LUCKY_DIMON [66]

The false statement from the above is that: Temporary charge imbalances in the molecules lead to London dispersion forces.

<h3>What are the factors that affect London dispersion forces?</h3>

Generally, the factors which affects the London dispersion forces a dispersion force are as follows:

Shape of the molecules
Distance between molecules
Polarizability of the molecules

However, London dispersion forces simply refers to a sort of temporary attractive force formed when electrons in two adjacent atoms occupy positions that make the atoms form dipoles.

So therefore, temporary charge imbalances in the molecules lead to London dispersion forces is a false statement

Learn more about London dispersion forces:

brainly.com/question/1454795

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2 years ago

HELP!! I WILL CROWN BRAINLEIST FOR FAST ANSWER! The element phosporus is located to the right of the element magnesium along the

Yuliya22 [10]

B. since Phosphorus has an atomic mass of 30.974 (larger) than Magnesium which is 24.305 (smaller)

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3 years ago

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