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Olenka [21]
3 years ago
10

True or False: The Equator receives the most DIRECT rays of the sun.

SAT
2 answers:
Ainat [17]3 years ago
8 0
True. Because sunlight arrives at a perpendicular (90 degree) angle to the Earth.
tensa zangetsu [6.8K]3 years ago
5 0
True because of the tilt of the Earth.
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Once alcohol has passed into the large intestine it is.
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it is quickly absorbed into the bloodstream

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Analyze how limitations to the right to freedom of expression of the media may assist in promoting public safety brainly
yuradex [85]

Limitations to the right to freedom of expression to the media, such as classified information, help in national security.

<h3>What are the Limitations to the Right to Freedom of Expression of the Media?</h3>

The media have special rights and duties to inform the society on issues that borders on the society for scrutiny, reflection or debate. However, freedom of speech and expression is not absolute as there are certain limitations to such rights a media can have in some cases.

Cases where freedom of expression of the media are restricted may include, sedition, slander, libel, classified information, or even incitement to an offense.

All these are punishable by the law in order to promote public safety and national security. So, these limitations help a lot in these ways.

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Find δs o for the combustion of methane to carbon dioxide and liquid water.
mote1985 [20]

The entropy change ΔS for the combustion of methane to carbon dioxide and liquid water is -242.6 J/mol.K

<h3>What is the Entropy in a Chemical system?</h3>

In a chemical system, Entropy (ΔS) is a quantitative measure of the variety of possible ways to disperse energy inside a chemical system.

  • In an isothermal process, entropy is determined as the quantity of heat transmitted to a system.

The chemical reaction for the combustion of methane can be expressed as:

\mathbf{CH_{4(g)} + 2O_{2(g)} \to CO_{2(g)} + 2H_2O_{(l)}}

The standard entropy of the combustion is expressed as:

\mathbf{\Delta S^0 =(S^0 _{CO_{2}} + 2 \times S^0 _{H_{2}O} )+(-S^0 _{CH_{4}}  -2 \times S^0 _{O_{2}} })

Replacing the standard entropies of the compounds into the above equation, we have:

\mathbf{\Delta S^0 =(213.8 \8 \ J/mol.K + 2 \times69.92 \ J/mol.K  )+(-186.16 \ J/mol.K -2 \times 205.04 \ J/mol.K})

\mathbf{\Delta S^0 =-242.6 \ J/mol.K}

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