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1 year ago

2C6H6 + 1502 →12CO2 + 6H20

Chemistry

1 answer:

solmaris [256]1 year ago

5 0

Answer:

.926 moles

Explanation:

Rounding :

H2 0 = 18 gm/mole

50 gm would then be 50 / 18 = 2.7777 moles of water

every two moles of 2 C6H6 produces 6 moles of water

2.7777/6 * 2 = .926 moles

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For the reaction: Cr2O7^2- + NH4^+ → Cr2O3 + N2, which is the oxidizing agent?

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Answer:

(NH4)2Cr2O7

Explanation:

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Answer:

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3 years ago

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A chemical reaction involves reactant species A, B, and C. Leaving all other factors identical, doubling the concentration of sp

umka21 [38]

Answer:

Rate = k . [B]² . [C]

Explanation:

The dependence of the reaction rate on the concentration of the reactants is given by the reaction order of each one, as shown in the rate equation.

$Rate=k.[A]^{x} .[B]^{y} .[C]^{z}$

where,

k is the rate constant

x, y, z are the reaction orders.

The rate of reaction is not affected by changing the concentration of species A. This means that the reaction order for A is x = 0 since when its concentration changes, the rate stays the same.

Leaving all other factors identical, doubling the concentration of species B increases the rate by a factor of 4. This means that the reaction order for B is y = 2, so when the concentration is doubled, the new rate is 2² = 4 times the initial rate.

The rate of the reaction is linearly dependent on the concentration of C. This means that the reaction order for C is z = 1, that is, a linear dependence.

All in all, the rate equation is:

Rate = k . [B]² . [C]

3 0

3 years ago

Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

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3 years ago