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Which of the following forms a molecular solid? Which of the following forms a molecular solid? C, graphite C10H22 CaO gold

Damm [24]

Answer:

C10H22

Explanation:

Molecular Solids comprises of a Vander waal's force of attraction between the molecule. These forces are very weak when compared to ionic and covalent bond.

In Carbon, Carbon is not a molecule but an atom. One of it unique characteristics is that it forms bonds with other carbon atoms. This property is know as catenation. The bond between these carbon atoms is know as covalent bond.

Graphite is an allotrope of carbon. It exists as black , slippery, hexagonal crystals.The carbon atoms in graphite forms flat layers and are joined together by strong covalent bonds. Graphite can be used as lubricant in engines.

Gold (Au) is an element on the periodic table with atomic number 79 and a mass number 197. It exists as a metal. Most times Gold forms hydrogen bonds.

C10H22 is known as decane. It is the tenth compound formed in the series of alkane family( an organic unsaturated carbon chain family). Alkanes are aliphatic hydrocarbons. The forces of attraction between the alkane family are weak.In decane , their exists Vander waal's force which makes Decane C10H22 a Molecular Solid.

6 0

3 years ago

If a system performs 147 kJ of work while receiving 47 kJ of heat, what is change in its internal energy?

Verdich [7]

Answer:

-100 kJ

Explanation:

We can solve this problem by applying the first law of thermodynamics, which states that:

$\Delta U = Q-W$

where:

$\Delta U$ is the change in internal energy of a system

Q is the heat absorbed/released by the system (it is positive if absorbed by the system, negative if released by the system)

W is the work done by the system (it is positive if done by the system, negative if done on the system)

For the system in this problem we have:

W = +147 kJ is the work done by the system

Q = +47 kJ is the heat absorbed by the system

So , its change in internal energy is:

$\Delta U = +47 - (+147) =-100 kJ$

6 0

2 years ago

Calculate the standard heat of reaction for the following methane-generating reaction of methanogenic bacteria: 4CH3NH2(g) + 2H2

PIT_PIT [208]

Answer: The standard heat for the given reaction is -138.82 kJ

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles.

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H_f_{(product)}]-\sum [n\times \Delta H_f_{(reactant)}]$

For the given chemical reaction:

$4CH_3NH_2(g)+2H_2O(l)\rightarrow 3CH_4(g)+CO_2(g)+4NH_3(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(3\times \Delta H_f_{(CH_4(g))})+(1\times \Delta H_f_{(CO_2(g))})+(4\times \Delta H_f_{(NH_3(g))})]-[(4\times \Delta H_f_{(CH_3NH_2(g))})+(2\times \Delta H_f_{(H_2O(l))})]$

We are given:

$\Delta H_f_{(H_2O(l))}=-285.8kJ/mol\\\Delta H_f_{(NH_3(g))}=-46.1kJ/mol\\\Delta H_f_{(CH_4(g))}=-74.8kJ/mol\\\Delta H_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H_f_{(CH_3NH_2(g))}=-22.97kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(3\times (-74.8))+(1\times (-393.5))+(4\times (-46.1))]-[(4\times (-22.97))+(2\times (-285.8))]\\\\\Delta H_{rxn}=-138.82kJ$

Hence, the standard heat for the given reaction is -138.82 kJ

3 0

2 years ago

A monoprotic weak acid when dissolved in water is 0.66% dissociated and produces a solution with a pH of 3.04. Calculate the Ka

raketka [301]

Answer:

Ka = 6.02x10⁻⁶

Explanation:

The equilibrium that takes place is:

HA ⇄ H⁺ + A⁻

Ka = [H⁺][A⁻]/[HA]

We calculate [H⁺] from the pH:

pH = -log[H⁺]

[H⁺] = $10^{-pH}$

[H⁺] = 9.12x10⁻⁴ M

Keep in mind that [H⁺]=[A⁻].

As for [HA], we know the acid is 0.66% dissociated, in other words:

[HA] * 0.66/100 = [H⁺]

We calculate [HA]:

[HA] = 0.138 M

Finally we calculate the Ka:

Ka = $\frac{[9.12x10^{-4}]*[9.12x10^{-4}]}{[0.138]}$ = 6.02x10⁻⁶

3 0

1 year ago

Pollen is transferred from the stamen of one flower to the stigma of another. What has occurred?

Digiron [165]

Cross-pollination is the transfer of pollen from the anther of one flower to the stigma of another flower on a different individual of the same species. Self-pollination occurs in flowers where the stamen and carpel mature at the same time, and are positioned so that the pollen can land on the flower's stigma.

3 0

2 years ago

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