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oksian1 [2.3K]
2 years ago
9

The reaction below shows how silver chloride can be synthesized. AgNO3 NaCl Right arrow. NaNO3 AgCl How many moles of silver chl

oride are produced from 15. 0 mol of silver nitrate? 1. 0 mol 15. 0 mol 30. 0 mol 45. 0 mol.
Chemistry
2 answers:
Otrada [13]2 years ago
7 0

From 15 mol of Silver nitrate, the moles of silver chloride produced have been 15 mol. Thus, option B is correct.

The balanced chemical reaction for the synthesis of silver chloride has been:

\rm AgNO_3\;+\;NaCl\;\rightarrow\;AgCl\;+\;NaNO_3

From the balanced equation, since there has presence equal moles of silver nitrate and sodium chloride, the moles of silver chloride formed has been equivalent. Thus, 1 mole of silver nitrate gives 1 mole of silver chloride.

The moles of silver nitrate available are,  \rm M_A_g_N_O_3=15\;mol

The moles of silver nitrate produced can be given as:

\rm 1\;mol\;AgNO_3=1\;mol\;AgCl\\15\;mol\;AgNO_3=15\;\times\;1\;mol\;AgCl\\15\;mol\;AgNO_3=15\;mol\;AgCl

Thus, the moles of silver chloride produced have been 15 mol. Thus, option B is correct.

For more information about moles produced, refer to the link:

brainly.com/question/10606802

hichkok12 [17]2 years ago
6 0

Answer:

15 mol

Explanation:

the balanced equation is AgNO3 + NaCl > NaNO3 + AgCl.

the stequiometry of the reactions says: one mol of silver nitrate reacts with one one mol of silver chloride to produce one mol of silver nitrate and one mol of sodium chloride.

by proportions, if you have 15 mol of AgNO3 you're gonna have 15 mol of AgCl

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Calculate the number of moles of an ideal gas if it occupies 1750 dm3 under 125,000 pa at a temperature of 127 c. a. 0.21 moles
labwork [276]

The number of mole will be 65.81 mole.

An ideal gas would be one for which both the overall volume of the molecules and even the forces that exist between them are so negligible as to have no influence on the behavior of something like the gas.

Number of ideal gas can be calculated by using the formula:

PV = nRT

where, p is pressure, n is number of mole, R is gas constant and T is temperature.

Given data:

V= 1750 dm^{3} = 1750 L

P = 125,000 p = 1.2 atm

R = 0.082 L /mole kelvin

T = 273+127 = 400 K

Now, put the value of given data in above equation.

1.23atm x 1750L = n x 0.0820atm x Liter/ mole x kelvin  x 400K

n = 65.81 mole.

Therefore, the number of mole will be 65.81 mole

To know more about mole

brainly.com/question/21050624

#SPJ4

6 0
2 years ago
Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
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Answer:

The entropy of the final solution decreases, as the reaction disorder is less.

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The higher the temperature, the greater the heat of the reaction and the greater the disorder it has, so the entropy will increase ... But this is not the case, since the solution cools, decreasing the entropy proportionally.

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