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3 years ago

What is the composition of a universal indicator?

Chemistry

1 answer:

Zina [86]3 years ago

6 0

Answer:

 The main universal indicator is in the form of a solution and are thymol blue , methyl red , bromothymol blue and phenolphthalein this mixture is important because each component loses or gains protons depending upon the acidity of the solution being tested .

maybe this mighr help u

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Calculate the heat released when 25.0 grams of water freezes at 0 degree Celsius.

mafiozo [28]

Answer:

= 8350 joules

Solution and explanation:

The heat of fusion refers to the quantity of heat released when a given amount of water freezes.

For example, 1 g of water releases 334 J when it freezes at 0°C.

Therefore; For 25.0 g of water.

Heat released = Mass of water × heat of fusion

= 25 g × 334 J/g

= 8350 Joules

Hence, the amount of heat released when 25.0 g of water freezes at 0°C is 8350 J.

5 0

3 years ago

Identify the element symbol 1s22s22p63s23p64s23d104p65s24d5

madam [21]

Answer:

Explanation:

You just have to follow the rows with the exponents. Just remember that when we get to d, the number in the front is a period lower. Hope this helps!

5 0

1 year ago

Cu(NO3)2 + Zn (s) → Cu (s) + Zn(NO3)2 is an example of which type of reaction?

labwork [276]

Single replacements since cu is being replaced by zn

4 0

4 years ago

Which is an example of a base?a(lemon juice b(baking soda c(an insolator d( water

bija089 [108]

Answer:

b) Baking soda

Explanation:

Baking soda is sodium hydrogen carbonate, NaHCO₃.

It reacts with water according to the equation

NaHCO₃ + H₂O ⇌ H₂CO₃ + NaOH

Thus, baking soda is a base because it increases the concentration of hydroxide ions in water.

Water is neutral.

Lemon juice is an acid because it contains citric acid.

An insulator is a substance that does not conduct electricity. Insulators are neither acidic nor basic.

3 0

3 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