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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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Please help thank you (15 points)
Eva8 [605]

Answer:

D. The sun has less gravitational pull on Pluto because it is farther away.

6 0
2 years ago
Titration of 0.824 g of potassium hydrogen phthalate required 38.314 g of naoh solution to reach the end point detected by pheno
melamori03 [73]

1.062 mol/kg.

<em>Step 1</em>. Write the balanced equation for the neutralization.

MM = 204.22 40.00

KHC8H4O4 + NaOH → KNaC8H4O4 + H2O

<em>Step 2</em>. Calculate the moles of potassium hydrogen phthalate (KHP)

Moles of KHP = 824 mg KHP × (1 mmol KHP/204.22 mg KHP)

= 4.035 mmol KHP

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 4.035 mmol KHP × (1 mmol NaOH/(1 mmol KHP)

= 4.035 mmol NaOH

<em>Step 4</em>. Calculate the mass of the NaOH

Mass of NaOH = 4.035 mmol NaOH × (40.00 mg NaOH/1 mmol NaOH)

= 161 mg NaOH

<em>Step 5</em>. Calculate the mass of the water

Mass of water = mass of solution – mass of NaOH = 38.134 g - 0.161 g

= 37.973 g

<em>Step 6</em>. Calculate the molal concentration of the NaOH

<em>b</em> = moles of NaOH/kg of water = 0.040 35 mol/0.037 973 kg = 1.062 mol/kg

3 0
3 years ago
A) The equilibrium shifts to the left, producing more H3O+.
Lunna [17]

Answer: these are just options, what is the main question, without it, we cannot determine which option is correct, so please repost the question.

5 0
3 years ago
The half-reaction occurring at the cathode in the balanced reaction shown below is __________?2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag
Shtirlitz [24]

Answer:

Correct choice are C and D (they are both, the same).

Explanation:

Cathode is the positive(+) electrode where a reduction occurs.

Reduction is the chemical reaction where the oxidation state is reduced.

2Ag(s) + 1/2 O2(g) + 2H+(aq) → 2Ag+(aq) + H2O (l)

A. 2H2O (l) → O2 (g) + 4H+ (aq) + 4e-

B. 2Ag (s) → 2Ag+ (aq) + 2e-

C. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

D. 1/2 O2 (g) + 2H+ (aq) + 2e- → H2O (l)

C or D, are ok. They are the same equation.

Oxygen from ground state reduce the oxidation state from 0 to -2

3 0
3 years ago
Calculate the frequency of radiation with a wavelength of 1.08 x 10^-6 m. simple explanation please​
pishuonlain [190]
C = vf
c stands for the speed of waves (which is a constant that is 3 x 10^8)
v stands for the wavelength (which is given)
f stands for frequency (what we are solving for)
3 x 10^8 = (1.08 x 10^-6)f
Divide both sides by the given wavelength
f = 2.78 * 10^14 seconds
8 0
3 years ago
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