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3 years ago

Lime is added to a lake to neutralize the effects of acid rain. The pH value of the

Chemistry

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

The change in concentration of H+ ions is -9.99×10⁻⁵

Explanation:

When pH rises, we talk about neutralization and, there is a decline in proton's concentration.

pH lower than 7 is acid

pH higher than 7 is basic

Protons are neutralized by hydroxides to make water as this:

H₃O⁺ + OH⁻ ⇄ H₂O

pH = - log [H₃O⁺]

[H₃O⁺] = 10^-pH

In conclussion: [H₃O⁺]₁ = 10⁻⁴ → 1×10⁻⁴

[H₃O⁺]₂ = 10⁻⁷ → 1×10⁻⁷

1×10⁻⁷- 1×10⁻⁴

The change in concentration of H+ ions is -9.99×10⁻⁵

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Which of these elements is unlikely to form covalent bonds? s, h, mg, ar, si?

kari74 [83]

<span>Answer: K because it is metal and typically forms ionic bonds. Ar is also unlikely to form any bonds because it has a full outer shell of electrons, but it can form covalent bonds.</span>

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3 years ago

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Which has more mass, a mole of water (H2O) or a mole of glucose (C6H12O6)? a. water b. glucose c. The masses are the same

riadik2000 [5.3K]

The correct answer is A

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3 years ago

Why were scientists unsure about the last common ancestor between apes and humans?

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Answer:

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3 years ago

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

kaheart [24]

The equilibrium constant of the reaction is 282. Option D

<h3>What is equilibrium constant?</h3>

The term equilibrium constant refers to the number that often depict how much the process is able to turn the reactants in to products. In other words, if the reactants are readily turned into products, then it follows that the equilibrium constant will be large and positive.

Concentration of bromine = 0.600 mol /1.000-L = 0.600 M

Concentration of iodine = 1.600 mol/1.000-L = 1.600M

In this case, we must set up the ICE table as shown;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -x -x +2x

E 0.6 - x 1.6 - x 1.190

If 2x = 1.190

x = 1.190/2

x = 0.595

The concentrations at equilibrium are;

[Br2] = 0.6 - 0.595 = 0.005

[I2] = 1.6 - 0.595 = 1.005

Hence;

Kc = [IBr]^2/[Br2] [I2]

Kc = ( 1.190)^2/(0.005) (1.005)

Kc = 282

Learn more about equilibrium constant:brainly.com/question/15118952

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2 years ago

If Kc = 4.0×10−2 for PCl3(g)+Cl2(g)⇌PCl5(g) at 520 K , what is the value of Kp for this reaction at this temperature?

Flauer [41]

Here we have to get the $K_{p}$ of the reaction at 520 K temperature.

The $K_{p}$ of the reaction is 1.705 atm

We know the relation between $K_{p}$ and $K_{c}$ is $K_{p}=K_{c}(RT)^{N}$ , where $K_{p}$ = The equilibrium constant of the reaction in terms of partial pressure, $K_{c}$ = The equilibrium constant of the reaction in terms of concentration and N = number of moles of gaseous products - Number of moles of gaseous reactants.

Now in this reaction, PCl₃ + Cl₂ ⇄ PCl₅

Thus number of moles of gaseous product is 1, and number of moles of gaseous reactants are 2. Thus N = |1 - 2| = 1 mole

The given value of $K_{c}$ is 4.0×10⁻²

The molar gas constant, R = 0.082 L. Atm. mol⁻¹. K⁻¹ and temperature, T = 520 K.

On plugging the values in the equation we get,

$K_{p} = 4.0 X 10^{-2}(0.082X520)^{1}$

Or, $K_{p}$ = 1.705 atm

Thus, the $K_{p}$ of the reaction is 1.705 atm

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3 years ago