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3 years ago

Please help with 2 chem questions thank you!

Chemistry

2 answers:

xeze [42]3 years ago

8 0

The correct answers are A and C.

andrew11 [14]3 years ago

6 0

The 1st one is A

The second one is I am 90% sure is A as well

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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⁻]

Where the concentrations [] are in equilibrium

Reaction quotient, Q, is defined as:

Q = [Ag⁺]² [CrO₄²⁻]

Where the concentrations [] are the actual concentrations

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

Where X is defined as the reaction coordinate

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

[CrO₄²⁻] = 0.135M

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3 years ago

I was born in ……… west. complete this sentense with article

Xelga [282]

Answer:

Article is

I was born in the  west .

4 0

2 years ago

Predict the products and then balance the equation

Aneli [31]

$\qquad \qquad\huge \underline{\boxed{\sf Answer}}$

Here's the balanced equation for given Double displacement reaction ~

$\sf Pb(NO_3)_2 +2 \: KI = PbI_2 +2 \: KNO_3$

The products fored are : Lead Iodide ( PbI2 ) and Potassium Nitrate ( KNO3 )

4 0

2 years ago

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A 25.00 ml sample of h3po4 solution is neutralized by exactly 54.93 ml of 0.04345 m ca(oh)2. what is the molarity of the h3po4 s

son4ous [18]

The neutralization equation is:
3 Ca(OH)₂ + 2 H₃PO₄ → Ca₃(PO₄)₂ + 6 H₂O

From this equation we can see that 3 moles of Ca(OH)₂ react with 2 moles of H₃PO₄

Numbers of mmol of Ca(OH)₂ = M x V = 0.04345 x 54.93 = 2.387 mmol

Number of mmol of H₃PO₄ = 2.387 x (2/3) = 1.591 mmol

Molarity of solution = n (mmol) / V(ml) = 1.591 / 25 = 0.0636 M

8 0

3 years ago

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When water freezes it's density gets lower the change in density is different than that of most substances most substances get d

icang [17]

When water is in liquid form its molecules are free to move around.
Water molecules are packed reasonably close together. However when water freezes its molecules take up a hexagonal lattice (repeating structure) which has space in the middle of it.

This is largely due to hydrogen bonding between water molecules (complicated).

As a result water molecules in ice aren't packed as closely together as they are in liquid water so the density of ice is lower than that of liquid water.

Hope that helps. I doubt you need to know about hydrogen bonding.

8 0

3 years ago