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2 years ago

In a reaction where A + B AB, what does an equilibrium position of A= 14%, B = 14%, and AB= 72% indicate about the reaction?

Chemistry

2 answers:

MissTica2 years ago

5 0

The answer is c I just answered mine .

fomenos2 years ago

3 0

<span>A.) Because A and B are equal, the reaction has reached equilibrium.
B.) There is more A and B than AB at equilibrium, and the reaction favors the reactants.
C.) There is more AB than A and B at equilibrium, and the reaction favors the products.
D.) There is more AB than A and B at equilibrium, and the reaction favors the reactants.
E.) The reaction is not at equilibrium.
These are the answer choices pls help
</span>

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Consider the bonds in iron(III) oxide. Are the bonds more ionic or more covalent?

kvv77 [185]

<span>the bonds in iron(III) oxide are more ionic</span>

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3 years ago

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(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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3 years ago

How many milliseconds are there in 3.5 seconds

azamat

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3 years ago

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What type of bond will form between nitrogen and bromine?

madreJ [45]

Answer:

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Explanation:

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2 years ago

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In the following reaction 2C6H6 + 15O2 12CO2 + 6H2O how many grams of oxygen will react with 10.47 grams of benzene (C6H6)?

IRINA_888 [86]

$2 C_{6}H_{6} + 15O_{2} ----->\ \textgreater \ 12CO_{2} + 6H_{2}O


$

$mol of benzene = \frac{mass}{Mr}$
= $\frac{10.47g}{(6 * 12) (6 * 1) g/mol}$
= 0.134 mol

mol of oxygen:
ratio of $C_{6} H_{6} : O_{2}$
= 2 : 15
= 1 : 7.5

: . mol of $O_{2}$ = 0.134mol * 7.5
= 1.01 mol

Mass of Oxygen = mol * Mr
= 1.01 mol * (16*2) g/mol
= 32.22g

Note: Mr is molar mass

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3 years ago