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Ganezh [65]
3 years ago
7

Select the correct answer.

Chemistry
2 answers:
Nuetrik [128]3 years ago
5 0

Answer:

A binary covalent compound is composed of two different elements (usually nonmetals). For example, a molecule of chlorine trifluoride, ClF3 contains 1 atom of chlorine and 3 atoms of fluorine.

Rule 1. The element with the lower group number is written first in the name; the element with the higher group number is written second in the name. Exception: when the compound contains oxygen and a halogen, the name of the halogen is the first word in the name.

Rule 2. If both elements are in the same group, the element with the higher period number is written first in the name.

Rule 3. The second element in the name is named as if it were an anion, i.e., by adding the suffix -ide to the root of the element name (e.g., fluorine = F, "fluoride" = F-; sulfur = S, "sulfide" = S2-).

Rule 4. Greek prefixes are used to indicate the number of atoms of each element in the chemical formula for the compound. Exception: if the compound contains one atom of the element that is written first in the name, the prefix "mono-" is not used.

Explanation:

V125BC [204]3 years ago
3 0

Answer:

the answer is D

Explanation:

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he rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy . If the rate c
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The question is incomplete, here is the complete question:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0 kJ/mol . If the rate constant of this reaction is 6.7 M^(-1)*s^(-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

<u>Answer:</u> The rate constant at 324°C is 61.29M^{-1}s^{-1}

<u>Explanation:</u>

To calculate rate constant at two different temperatures of the reaction, we use Arrhenius equation, which is:

\ln(\frac{K_{324^oC}}{K_{244^oC}})=\frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_{244^oC} = equilibrium constant at 244°C = 6.7M^{-1}s^{-1}

K_{324^oC} = equilibrium constant at 324°C = ?

E_a = Activation energy = 71.0 kJ/mol = 71000 J/mol   (Conversion factor:  1 kJ = 1000 J)

R = Gas constant = 8.314 J/mol K

T_1 = initial temperature = 244^oC=[273+244]K=517K

T_2 = final temperature = 324^oC=[273+324]K=597K

Putting values in above equation, we get:

\ln(\frac{K_{324^oC}}{6.7})=\frac{71000J}{8.314J/mol.K}[\frac{1}{517}-\frac{1}{597}]\\\\K_{324^oC}=61.29M^{-1}s^{-1}

Hence, the rate constant at 324°C is 61.29M^{-1}s^{-1}

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