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Ulleksa [173]
3 years ago
5

This equation is not balanced. Click on the coefficient we should change first. help ASAP

Chemistry
1 answer:
horrorfan [7]3 years ago
7 0

Answer:  The coefficient we should change first is for NO_{3}.

Explanation:

The given reaction equation is as follows.

Al + Ni(NO_{3})_{2} \rightarrow Al(NO_{3})_{3} + Ni

Here, number of atoms present on reactant side are as follows.

  • Al = 1
  • Ni = 1
  • NO_{3} = 2

Number of atoms present on the product side are as follows.

  • Al = 1
  • Ni = 1
  • NO_{3} = 3

To balance this equation, multiply Al by 2 and NO_{3} by 3 on reactant side. Also, multiply Al(NO_{3})_{3} by 2 and Ni by 3 on the product side.

Hence, the equation can be rewritten as follows.

2Al + 3Ni(NO_{3})_{2} \rightarrow 2Al(NO_{3})_{3} + 3Ni

Now, number of atoms present on reactant side are as follows.

  • Al = 2
  • Ni = 3
  • NO_{3} = 6

Number of atoms present on product side are as follows.

  • Al = 2
  • Ni = 3
  • NO_{3} = 6

Since, the atoms on both reactant and product side are same. Hence, it is now a balanced chemical equation.

Thus, we can conclude that the coefficient we should change first is for NO_{3}.

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A 265-mL flask contains pure helium at a pressure of 751 torrs. A second flask with a volume of 465 mL contains pure argon at a
Nadya [2.5K]

Answer:

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

Explanation:

Step 1: Data given

Volume of the flask helium = 265 mL

Pressure in the helium flask = 751 torr = 751/760 atm

Volume of the flask argon = 465 mL

Pressure in the argon flask = 727 torr = 727/760 atm

The total pressure exerted by a gaseous mixture is equal to the sum of the partial pressures of each individual component in a gas mixture.

Step 2: Calculate total volume

Total volume = 265 mL + 465 mL = 730 mL =  0.730 L

Step 3: Boyle's Law:

P1V1=P2V2

⇒ with P1 = total pressure gas exerts in its own flask

 ⇒ with V1 = volume of flask with stopcock valve closed

 ⇒ with P2 = partial pressure of gas exerts on total volume of both flasks when stopcock valve is opened  

 ⇒ with V2 = total volume of both flasks with stopcock valve opened

Helium using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of helium = 751 /760 = 0.98816 atm

 ⇒ with V1 = volume of helium = 0.265 L

 ⇒ with P2 = The new partial pressure of helium

 ⇒ with V2 = total volume = 0.730 L

(0.98816 atm)(0.265L)=P2(0.730L)

P2=0.359 atm

Argon using Boyle's Law equation from above:

P1V1=P2V2

⇒ with P1 = Pressure of argon = 727/760 = 0.95658 atm

 ⇒ with V1 = volume of argon = 0.465 L

 ⇒ with P2 = The new partial pressure of argon

 ⇒ with V2 = total volume = 0.730 L

(0.95658 atm)(0.465L)=P2(0.730L)

P2=0.609 atm

Step 4: Convert pressure in atm to torr

Pressure helium = 0.359 atm = 272.8 torr

Pressure argon = 0.609 atm = 472.8 torr

Step 5: Calculate Total pressure

Ptotal = P(He)+P(Ar)

⇒ Pt  = total pressure of the gas mixture

⇒ P(He) = partial pressure of Helium

 ⇒ P(Ar)  = partial pressure of Argon

Pt = 272.8 torr + 472.8 torr

Pt = 745.6 torr

Total Pressure = 745.6 torr

Partial Pressure of He = 272.8 torr

Partial Pressure of Ar =  472.8 torr

5 0
3 years ago
jenny dissolves as much salt as she can in 100mL of room temperature water. she learned in class that less salt will dissolve in
Slav-nsk [51]

Answer:

if ice is added more salt can be dissolved ...........

5 0
3 years ago
Read 2 more answers
How many moles are in 12.5 g ammonium chloride
77julia77 [94]

Answer:

0.233682161601125

Explanation:

3 0
3 years ago
I need help! how many moles are in 3.4*10^-7 grams of silicon dioxide, SiO2. I got 2.0^17 but I think it's wrong :(
blondinia [14]
Molar mass SiO2 = 28 + 32 = 60 

<span>so moles sand = 3.4 x 10-7 / 60</span>
8 0
3 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
4 0
3 years ago
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