The ease which an acid or base dissociates in a solution refers to

Consider the following equilibrium: B(aq)+H2O(l)⇌HB+(aq)+OH−(aq) Suppose that a salt of HB+ is added to a solution of B at equil

Liula [17]

Answer:

Part A: C. The equilibrium constant for the reaction will stay the same.

Part B: A. The concentration of B(aq) will increase.

Part C: C. The pH of the solution will decrease.

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

Suppose that a salt of HB+ is added to a solution of B at equilibrium.

Part A Will the equilibrium constant for the reaction increase, decrease, or stay the same?

Adding HB+ will increase the concentration of the products side. so, the reaction will be shifted to the lift direction to suppress the effect of adding HB+ and attain the equilibrium again.

so, The equilibrium constant for the reaction will stay the same.

Part B Will the concentration of increase, decrease, or stay the same?

 Adding HB+ will increase the concentration of the products side. so, the reaction will be shifted to the lift direction to suppress the effect of adding HB+ and attain the equilibrium again.

So, the concentration of species in the reactant sides will increase.

Thus, the right choice is:

A. The concentration of B(aq) will increase.

Part C Will the pH of the solution increase, decrease, or stay the same?

 Adding HB+ will increase the concentration of the products side. so, the reaction will be shifted to the lift direction to suppress the effect of adding HB+ and attain the equilibrium again.

So, HB+ reacts with a large amount of OH- to attain the equilibrium again and shifting the reaction towards the lift side to attain the equilibrium again.

Which means decreasing the concentration of OH-.

As the concentration of OH- decreased, the pH of the solution decreased.

So, the right choice is:

c. The pH of the solution will decrease.

6 0

3 years ago

Determine the amount of energy (heat) in joules required to raise the temperature of 5 kg of iron from 50°C to 200°C.

larisa [96]

333,000 Joules is the amount of energy (heat) in joules required to raise the temperature of 5 kg of iron from 50°C to 200°C.

Explanation:

Data given:

mass of iron 5 Kg or 5000

initial temperature = 50 degree centigrade

final temperature = 200 degrees centigrade

change in temperature (Δ T= 200 -50 degrees centigrade)

= 150 °C

cp (specific heat capacity of iron) = 0.444j/gram C

q (heat supplied) = ?

applying the formula,

q=mcΔT

putting the values in the equation:

q = 5000 X 0.444 X150

q = 333,000 Joules of energy.

The heat required 333,000 Joules of energy is required.

4 0

4 years ago

What mass of phosphorus and of chlorine is needed to produce 87 g of pentachloride if the reaction yield is 60%?

Anna11 [10]

Hey there!

theoretical yield = (87* 100 ) / 60 = 145 g of phosphorous pentachloride

the reaction for this process is :

molar mass P2O5 => 239.2125 g/mol

molar mass P4 => 123.90 g/mol

molar mass Cl2 => 70.9060 g/mol

P4 + 5 Cl2 --------------> 2 P2Cl5

moles of P2O5 :

145 g / 239.2125 => 0.6062 moles of P2O5

Therefore:

moles of P4 = 0.6062 mol / 2 => 0.3031 moles

moles of Cl2 = 5/2 * 0.6062 mol = 1.5154 moles

mass of P4 ( phosphours ) = 0.3031 mol * 123.90 => 37.554 grams

mass of Cl2 ( chlorine ) = 1.5154 mol * 70.9060 => 107.45 grams

Hope That helps!

3 0

3 years ago

Stoichiometry Problems!

lisov135 [29]

Hey there!

C₆H₁₂O₆(s) + 6O₂(g) => 6CO₂(g) + 6H₂0(l)

a.)

First we need to find how many molecules of oxygen gas we need.

Every one molecule of C₆H₁₂O₆ will react with six molecules of O₂. So, if we have 3.011 x 10²³ molecules of C₆H₁₂O₆, we need six times that of oxygen.

3.011 x 10²³ x 6 = 18.066 x 10²³ = 1.8066 x 10²⁴

So we need 1.8066 x 10²⁴ molecules of O₂. We need to find the volume of this in liters.

At STP, one mole of a gas occupies 22.4 liters. Let's find the number of moles we have of O₂.

(1.8066 x 10²⁴) ÷ (6.022 x 10²³) = 3 moles

3 x 22.4 = 67.2

67.2 liters of O₂ is needed.

b.)

Okay, so to find the percent yield, we need to find the theoretical yield and the actual yield. We are given the actual yield, so what we need is the theoretical yield.

For every one mole of C₆H₁₂O₆, theoretically 6 moles of H₂O will be produced.

Let's convert grams to moles for C₆H₁₂O₆:

1 gram / 180 grams = 0.0055556 moles C₆H₁₂O₆

Theoretically, 6 times that is the moles of H₂O produced:

0.0055556 x 6 = 0.033333 moles H₂O

Molar mass of H₂O is 18.015, so let's find grams:

0.033333 x 18.015 = 0.600 grams H₂O

So we have our theoretical yield, 0.600, and our actual yield, 0.303.

0.303 ÷ 0.600 = 0.505

Convert to a percent: 0.505 x 100 = 50.5%

The percent yield is 50.5%.

Hope this helps!

4 0

3 years ago

The initial volume of HCl was 1.25 ml and LiOH was 2.65 ml. The final volume of HCL was 13.60 ml and LiOH was 11.20 ml. If the L

Dennis_Churaev [7]

Q1)
This is a strong acid- strong base base reaction, balanced equation for the reaction is as follows;
LiOH + HCl ---> LiCl + H₂O
stoichiometry of acid to base is 1:1
volume of HCl used up - 13.60 - 1.25 = 12.35 mL
volume of LiOH used up - 11.20 - 2.65 = 8.55 mL
molarity of LiOH - 0.140 M
The number of LiOH moles reacted - $\frac{0.140 mol/L*8.55mL}{1000mL}$ = 0.001197 mol
according to stoichiometry, number of LiOH moles = number of HCl moles
Therefore number of HCl moles reacted - 0.001197 mol
The number of HCl moles in 12.35 mL - 0.001197 mol
Then number of HCl moles in 1000 mL - $\frac{0.001197*1000mL}{12.35mL}$
Molarity of HCl - 0.0969 M

Q2)
Volume of HCl used - 12.35 mL
Volume of LiOH used - 8.55 mL
Molarity of HCl - 0.140 M
In 1 L solution of HCl there are 0.140 mol of HCl
Therefore number of HCl moles in 12.35 mL - $\frac{0.140mol*12.35 mL}{1000mL}$
Number of HCl moles reacted - 0.001729 mol
since molar ratio of acid to base is 1:1
the number of LiOH moles that reacted - 0.001729 mol
Therefore number of moles in 8.55 mL - 0.001729
Then number of LiOH moles in 1000 mL - $\frac{0.001729mol*1000mL}{8.55 mL}$
molarity of LiOH - 0.202 M

5 0

3 years ago