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2 years ago

10

Within a particular compound, an atom gives up two electrons. This atom is

2 answers:

ValentinkaMS [17]2 years ago

6 0

Answer:

The correct answer is option B. positively charged.

Explanation:

A positively charged ion is an atom that readily loses its one or more electrons in a compound. Positively charged ion generally has same number of electrons and protons which means they are neutral charge atom. These atoms are generally attracted to negatively charged ion that lack electron in outer shell.

Thus, the correct answer is option. B. positively charged.

lys-0071 [83]2 years ago

3 0

It is a positively charges because when an atom gives up electrons it becomes positively charges. Think of it like this.....its being generous so it deserves to be a positive person. ;D.
It cant be A. because a covalent bond only shares and doesn't give up electrons. Our answer must be an ion.
Its not negatively charged because it didn't gain any.

So our answer is B. positively charged
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Explanation:

According to the Arrhenius equation,

$K=A\times e^{\frac{-Ea}{RT}}$

or,

$\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$K_1$ = rate constant at $T_1$ = 1.00

$K_2$ = rate constant at $T_2$ = 5.00

$Ea$ = activation energy for the reaction = 28.90 kJ/mol= 28900 j/mol

R = gas constant = 8.314 J/mole.K

$T_1$ = initial temperature = 313 K

$T_2$ = final temperature = ?

Now put all the given values in this formula, we get

$\log (\frac{5.00}{1.00})=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]$

$0.69=\frac{28900}{2.303\times 8.314J/mole.K}[\frac{1}{313K}-\frac{1}{T_2K}]$

$T_2=365K$

Therefore, 365 K is required to increase the reaction rate by 5.00 times.

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