All categories

3 years ago

Describe how you could show that an alkaline solution is formed when caesium reacts with water

1 answer:

3 0

After reacting, place a red moist litmus paper into the final solution and the litmus paper will turn blue. This indicates that the solution is alkaline.
Alternative methods
Using universal indicator
using phenophtalein
using bromothyl blue
using methyl orange
use pH probe connected to a data logger

You might be interested in

What chemical substance is formed when the oh and h is joined?

Nezavi [6.7K]

Water is formed when this happens

3 0

3 years ago

Depending on the reaction, we could monitor the progress towards equilibrium by observing __________.

Alex Ar [27]

Depending on the reaction, we could monitor the progress towards equilibrium by observing the concentration of the reactant and the product are equal with time.

<h3>What is equilibrium?</h3>

Equilibrium is a stage of reaction in which the rate of forwarding reaction is equal to the rate of backward reaction and equilibrium is stable at the reversible state of mode.

The concentration of reactant and product must also be equal or the same as the time then only it can be an equilibrium reaction.

Therefore equilibrium depends on the reaction, the concentration of the reactant and the product are equal with time.

Learn more about equilibrium, here:

brainly.com/question/13463225

#SPJ4

4 0

2 years ago

The boiling point and distillation temperature of a substance are the same. true false

Novosadov [1.4K]

The boiling point and distillation temperature of a substance are the same. The answer would be True

3 0

3 years ago

The temperature and number of moles of a gas are held constant. Which of the following is true for the pressure of the gas?

Taya2010 [7]

Answer:

Pressure is inversely proportional to the volume of gas.

Explanation:

According to Boyle's law,

The volume of given amount of gas is inversely proportional to the pressure applied on gas at constant volume and number of moles of gas.

Mathematical expression:

P ∝ 1/ V

P = K/V

PV = K

when volume is changed from V1 to V2 and pressure from P1 to P2 then expression will be.

P1V1 = K P2V2 = K

P1V1 = P2V2

3 0

4 years ago

Calculate the activity coefficients for the following conditions:

uysha [10]

Answer:

For a: The activity coefficient of copper ions is 0.676

For b: The activity coefficient of potassium ions is 0.851

For c: The activity coefficient of potassium ions is 0.794

Explanation:

To calculate the activity coefficient of an ion, we use the equation given by Debye and Huckel, which is:

$-\log\gamma_i=\frac{0.51\times Z_i^2\times \sqrt{\mu}}{1+(3.3\times \alpha _i\times \sqrt{\mu})}$ ........(1)

where,

$\gamma_i$ = activity coefficient of ion

$Z_i$ = charge of the ion

$\mu$ = ionic strength of solution

$\alpha _i$ = diameter of the ion in nm

To calculate the ionic strength, we use the equation:

$\mu=\frac{1}{2}\sum_{i=1}^n(C_iZ_i^2)$ ......(2)

where,

$C_i$ = concentration of i-th ions

$Z_i$ = charge of i-th ions

For a:

We are given:

0.01 M NaCl solution:

Calculating the ionic strength by using equation 2:

$C_{Na^+}=0.01M\\Z_{Na^+}=+1\\C_{Cl^-}=0.01M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.01\times (+1)^2)+(0.01\times (-1)^2)]\\\\\mu=0.01M$

Now, calculating the activity coefficient of $Cu^{2+}$ ion in the solution by using equation 1:

$Z_{Cu^{2+}}=2+\\\alpha_{Cu^{2+}}=0.6\text{ (known)}\\\mu=0.01M$

Putting values in equation 1, we get:

$-\log\gamma_{Cu^{2+}}=\frac{0.51\times (+2)^2\times \sqrt{0.01}}{1+(3.3\times 0.6\times \sqrt{0.01})}\\\\-\log\gamma_{Cu^{2+}}=0.17\\\\\gamma_{Cu^{2+}}=10^{-0.17}\\\\\gamma_{Cu^{2+}}=0.676$

Hence, the activity coefficient of copper ions is 0.676

For b:

We are given:

0.025 M HCl solution:

Calculating the ionic strength by using equation 2:

$C_{H^+}=0.025M\\Z_{H^+}=+1\\C_{Cl^-}=0.025M\\Z_{Cl^-}=-1$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.025\times (+1)^2)+(0.025\times (-1)^2)]\\\\\mu=0.025M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.025M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.025}}{1+(3.3\times 0.3\times \sqrt{0.025})}\\\\-\log\gamma_{K^{+}}=0.070\\\\\gamma_{K^{+}}=10^{-0.070}\\\\\gamma_{K^{+}}=0.851$

Hence, the activity coefficient of potassium ions is 0.851

For c:

We are given:

0.02 M $K_2SO_4$ solution:

Calculating the ionic strength by using equation 2:

$C_{K^+}=(2\times 0.02)=0.04M\\Z_{K^+}=+1\\C_{SO_4^{2-}}=0.02M\\Z_{SO_4^{2-}}=-2$

Putting values in equation 2, we get:

$\mu=\frac{1}{2}[(0.04\times (+1)^2)+(0.02\times (-2)^2)]\\\\\mu=0.06M$

Now, calculating the activity coefficient of $K^{+}$ ion in the solution by using equation 1:

$Z_{K^{+}}=+1\\\alpha_{K^{+}}=0.3\text{ (known)}\\\mu=0.06M$

Putting values in equation 1, we get:

$-\log\gamma_{K^{+}}=\frac{0.51\times (+1)^2\times \sqrt{0.06}}{1+(3.3\times 0.3\times \sqrt{0.06})}\\\\-\log\gamma_{K^{+}}=0.1\\\\\gamma_{K^{+}}=10^{-0.1}\\\\\gamma_{K^{+}}=0.794$

Hence, the activity coefficient of potassium ions is 0.794

6 0

3 years ago