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natima [27]
2 years ago
11

Complete and balance the molecular equation for the reaction of aqueous sodium carbonate, Na 2 CO 3 Na2CO3 , and aqueous nickel(

II) chloride, NiCl 2 NiCl2 . Include physical states. molecular equation:
Na 2 CO 3 ( aq ) + NiCl 2 ( aq ) ⟶ 2 NaCl ( aq ) + NiCO 3 ( s ) Na2CO3(aq)+NiCl2(aq)⟶2NaCl(aq)+NiCO3(s) Enter the balanced net ionic equation for this reaction. Include physical states. net ionic equation:
Chemistry
1 answer:
antoniya [11.8K]2 years ago
3 0

<u>Answer:</u> The net ionic equation is written below.

<u>Explanation:</u>

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of sodium carbonate and nickel (II) chloride is given as:

Na_2CO_3(aq.)+NiCl_2(aq.)\rightarrow 2NaCl(aq.)+NiCO_3(s)

Ionic form of the above equation follows:

2Na^{+}(aq.)+CO_3^{2-}(aq.)+Ni^{2+}(aq.)+2Cl^{-}(aq.)\rightarrow NiCO_3(s)+2Na^+(aq.)+2Cl^-(aq.)

As, sodium and chloride ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

Ni^{2+}(aq.)+CO_3^{2-}(aq.)\rightarrow NiCO_3(s)

Hence, the net ionic equation is written above.

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The chlorination of methane occurs in a number of steps that results in the formation of chloromethane and hydrogen chloride. Th
kenny6666 [7]

Answer:

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

Explanation:

For the reaction:

2CH₄(g)+3Cl₂(g)⟶2CH₃Cl(g)+2HCl(g)+2Cl⁻(g)

295 mL≡ 0,295L of methane at STP are:

n = PV/RT

P = 1 atm; V = 0,295L; R = 0,082atmL/molK; T = 273K.

moles of methane: 0,0132 moles

For 725 mL of chlorine ≡ 0,725L

moles of chlorine at STP are: ≡ 0,0324 moles

For a complete reaction of 0,0132 moles of CH₄:

0,0132 mol CH₄× \frac{3molCl_{2}}{2 molCH_{4}} = <em>0,0198 moles</em>

The reaction reaches 77%, moles of Cl₂ that react are: 0,0198×77% = 0,0153 mol

As you have 0,0324 moles of Cl₂, moles that will not react are:

0,0324 - 0,0153 = <em>0,0171 mol Cl₂</em>

As the reaction reaches 77% completion, moles of CH₄ that react are:

0,0132×77% =<em> 0,0102 moles of CH₄ And the moles that don't react are </em><em>0,00300 mol</em>

Thus, moles of each compound are:

0,0102 moles of CH₄×\frac{3Cl_{2}}{2 molCH_{4}}= <em>0,0153 mol  + 0,0171 mol = 0,0324 mol Cl₂</em>

0,0102 moles of CH₄×\frac{2CH_{3}Cl}{2 molCH_{4}}= <em>0,0102 mol CH₄</em>

0,0102 moles of CH₄×\frac{2HCl}{2 molCH_{4}}= <em>0,0102 mol HCl</em>

0,0102 moles of CH₄×\frac{2Cl^{-}}{2 molCH_{4}}= <em>0,0102 mol Cl⁻</em>

Total pressure using:

P = nRT/V

Where: n = 0,0102mol×3+0,0324mol + 0,0030mol = 0,0660mol; R = 0,082 atmL/molK; T = 298K; V = 2L

Total pressure = 0,806 atm

Partial pressure of CH₄: 0,037 atm

Partial pressure of Cl₂: 0,396 atm

Partial pressure of CH₃Cl: 0,125 atm

Partial pressure of HCl: 0,125 atm

Partial pressure of Cl⁻: 0,125 atm

<em>-To obtain partial pressure you change the moles for each compound-</em>

<em />

I hope it helps!

5 0
2 years ago
A line in the Brackett series of hydrogen has a wavelength of 1945 nm. From what state did the electron originate?
fiasKO [112]
Rydberg Eqn is given as:
 1/λ = R [1/n1^2 - 1/n2^2] 
<span>Where λ is the wavelength of the light; 2626 nm = 2.626×10^-6 m </span>
<span>R is the Rydberg constant: R = 1.09737×10^7 m-1 </span>
<span>From Brackett series n1 = 4 </span>
<span>Hence 1/(2.626×10^-6 ) = 1.09737× 10^7 [1/4^2 – 1/n2^2] </span>
<span>Some rearranging and collecting up terms: </span>
<span>1 = (2.626×10^-6)×(1.09737× 10^7)[1/16 -1/n2^2] </span>
<span>1= 28.82[1/16 – 1/n2^2] </span>
<span>28.82/n^2 = 1.8011 – 1 = 0.8011 </span>
<span>n^2 = 28.82/0.8011 = 35.98 </span>
<span>n = √(35.98) = 6</span>
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The reaction of benzene with (CH3)3CCH2Cl in the presence of anhydrous aluminum chloride produces principally which of these?
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Answer:

See explanation and image attached

Explanation:

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