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Soloha48 [4]
2 years ago
13

HELP! URGENTLY NEEDED! I NEED THIS ANSWER IN 5 MINUTES AND PERSON WHO HELPS GET THE ANSWER RIGHT GETS 25 POINTS!!! PLEASE I AM F

AILING THIS CLASS AND NEED SO MUCH HELP!!!!
The coefficient/s will balance the chemical equation CH4 +O2 -------» CO2 + H2O is


a

1, 2, 1, 2

b

2, 1, 1, 2

c

2, 2, 1, 1

d

1, 1, 1, 2
Chemistry
2 answers:
marishachu [46]2 years ago
7 0

Answer:

hey miss

Explanation:

gtnhenbr [62]2 years ago
7 0

The coefficients : a. 1,2,1,2

<h3>Further explanation</h3>

Given

Reaction(unbalanced)

CH4 +O2 -------» CO2 + H2O

Required

The blanced equation

Solution

Give a coefficient

CH4 +aO2 -------» bCO2 + cH2O

Make an equation

C, left=1, right=b⇒b=1

H, left=4, right = 2c⇒2c=4⇒c=2

O, left=2a, right=2b+c⇒2a=2b+c⇒2a=2.1+2⇒a=2

The equation becomes :

CH4 +2O2 -------» CO2 + 2H2O

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Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na 2CrO 4 are added to 250 mL of 0.12 M A
olchik [2.2K]

Answer:

A precipitate will form.

[Ag⁺] = 2.8x10⁻⁵M

Explanation:

When Ag⁺ and CrO₄²⁻ are in solution, Ag₂CrO₄(s) is produced thus:

Ag₂CrO₄(s) ⇄ 2 Ag⁺(aq) + CrO₄²⁻(aq)

Ksp is defined as:

Ksp = 1.1x10⁻¹² = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are in equilibrium</em>

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

<em>Where the concentrations [] are the actual concentrations</em>

<em />

If Q < Ksp, no precipitate will form, if Q >= Ksp, a precipitate will form,

The actual concentrations are -Where 500mL is the total volume of the solution-:

[Ag⁺] = [AgNO₃] = 0.12M ₓ (250mL / 500mL) = 0.06M

[CrO₄²⁻] = [Na₂CrO₄] = 0.33M × (250mL / 500mL) = 0.165M

And Q = [0.06M]² [0.165M] = 5.94x10⁻⁴

As Q > Ksp; a precipitate will form

In equilibrium, some Ag⁺ and some CrO₄⁻ reacts decreasing its concentration until the system reaches equilibrium. Equilibrium concentrations will be:

[Ag⁺] = 0.06M - 2X

[CrO₄²⁻] = 0.165M - X

<em>Where X is defined as the reaction coordinate</em>

<em />

Replacing in Ksp expression:

1.1x10⁻¹² = [0.06M - 2X]² [0.165M - X]

Solving for X:

X = 0.165M → False solution. Produce negative concentrations.

X = 0.0299986M

Replacing, equilibrium concentrations are:

[Ag⁺] = 0.06M - 2(0.0299986M)

[CrO₄²⁻] = 0.165M - 0.0299986M

<h3>[Ag⁺] = 2.8x10⁻⁵M</h3>

[CrO₄²⁻] = 0.135M

6 0
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