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Mkey [24]
3 years ago
14

Write an equation that shows the formation of a chromium(ii) ion from a neutral chromium atom.

Chemistry
2 answers:
daser333 [38]3 years ago
8 0
The   equation  that  shows  the  formation  of  chromium (ii) ion  from  neutral  chromium  atom  is  as  follow
Cr ---> cr^2+   +  2e-
Cr^2+ is  the  chromium  ion   with  oxidation  state  of  two  which  is  one  of  the  common  ion  of  chromium.  Other  common  ion   of  chromium  include  chromium  of  oxidation  state  6  and  3

dangina [55]3 years ago
5 0

Answer : The balanced equation will be,

Cr\rightarrow Cr^{2+}+2e^-

Explanation :

Oxidation : It is the process of losing of one or more electrons.

Reduction : It is the process of gaining of one or more electrons.

The balanced equation will be,

Cr\rightarrow Cr^{2+}+2e^-

A neutral chromium atom is oxidized into chromium(II) ions by the loss of two electrons.


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How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?
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Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

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c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

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