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Paladinen [302]
3 years ago
12

Consider the chemical equation. 2NBr3 + 3NaOH mc008-1.jpg N2 + 3NaBr + 3HOBr If there are 40 mol of NBr3 and 48 mol of NaOH, wha

t is the excess reactant?
Chemistry
2 answers:
Vladimir79 [104]3 years ago
6 0
The correct answer o your amazing question is NBr3
Anna [14]3 years ago
3 0

Answer is: NBr₃ is the excess reactant.

Balanced chemical reaction: 2NBr₃ + 3NaOH → N₂ + 3NaBr + 3HBrO.

n(NBr₃) = 40 mol; amount of substance.

n(NaOH) = 48 mol.

From chemical reaction: n(NBr₃) : n(NaOH) = 2 : 3.

40 mol : n(NaOH) = 2 : 3.

n(NaOH) = 60 mol; for 40 moles of NBr₃, 60 moles of NaOH is needed, because there is only 48 moles, not all NBr₃ will react (onl 32 moles, 16 moles are excess).

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A 0.1000 mol L^-1 aqueous solution of HCl in a cell with plates having area 7.200 cm² separated by 3.600 cm has a resistance R =
Alex17521 [72]

Explanation:

The given data is as follows.

          Molarity = 0.1 M,             Area = 7.2 cm^{2}

          Resistance = 12.553 ohm,      Length = 3.6 cm

As it is known that relation between resistance, length and area is as follows.

                    R = \rho \frac{l}{A}

and,       \frac{1}{R} = \frac{1}{\rho} \times \frac{A}{l}  

                      k = c \times x

where,     k = specific conductivity

               c = conductance

                x = cell constant

Therefore, value of c = \frac{1}{R} = \frac{1}{12.553} = 0.0796 per ohm

             x = \frac{l}{A} = \frac{3.6 cm}{7.2 cm^{2}}

                           = 0.5 per cm

Hence, calculate the value of specific conductivity as follows.

                 k = c \times x

                          = 0.0796 per ohm \times 0.5 per cm

                          = 0.0398 per ohm per cm

Relation between molar  ionic conductivity and specific conductivity is as follows.

             \lambda_{m} = \frac{k \times 1000}{M}

                          = \frac{0.0398 \text{per ohm per cm} \times 1000}{0.1 M}

                          = 398 \Omega^{-1} cm^{2} mol^{-1}

Also, \Omega^{-1} = Siemen

                \lambda_{m} = 398 S cm^{2} mol^{-1}

thus, we can conclude that value of molar ionic conductivity of given hydrogen ions is 398 S cm^{2} mol^{-1}.

6 0
3 years ago
If a compound contains an anion which is the conjugate base of a weak acid, the addition of hydronium ion solubility does what ?
wel

Answer:

The addition of hydronium ion increases the solubility of the salt

Explanation:

When a weak acid, MA dissociates in an aqueous solution, the following products in equilibrium are obtained;

MA ----> M+ + A-

The anion which is a conjugate base will be removed from the solution with the addition of a hydronium ion from an acid: 2H3O+ + 2A- ----> 2HA + 2H20

This distorts the reaction equilibrium. According to Le Chatelier’s principle, more MA will dissolve until in order to restore the previous equilibrium of the reaction.

Therefore, an acidic pH increases the solubility of almost all sparingly soluble salts whose anion is the conjugate base of a weak acid.

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4 years ago
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gogolik [260]

Answer:

-0.044 V

Explanation:

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8 0
3 years ago
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