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4 years ago

What type of bond is H2Se

1 answer:

6 0

Answer:The lone pair of electrons takes up more space than a regular bonding pair since it it is not confined to be between two atoms, so it adds coulombic repulsion to the bonding pairs and compresses the angle. Therefore, the bond angle is less than the standard 109.5∘ . It is actually 97.7∘

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Small, highly charged metal cations may interact with H2O to give solutions that are not neutral. Which net ionic chemical equat

Vaselesa [24]

Answer:

Option B is correct.

The net ionic chemical reaction when Cu(NO₃)₂ reacts with water is

Cu²⁺(aq) + H₂O(l) ⇌ Cu(OH)¹⁺(aq) + H¹⁺(aq)

Explanation:

When ions dissolve in polar solvents (in water especially), then often attract ions of opposite signs.

Cu(NO₃)₂ dissolves into Cu²⁺ & NO₃⁻ and water contains H¹⁺ & OH⁻

The Cu²⁺ attracts the OH⁻ from water, thereby giving the Cu(OH)¹⁺ ion (with a net charge of +2-1 = +1) and the NO₃⁻ takes up the H⁺ ion.

8 0

3 years ago

How to make 0.01 Molar solution in 30 ml of water

Leviafan [203]

Answer:

You see the concentration decreased ten fold, from 1 M to 0.1 M.

You can simply increase the volume by ten times.

For example

If you have 250 mL of 1 molar HCl, you can add distilled water upto 2500 mL. Now the concentration is 0.1 molar.

If you want to use the serial dilution method, you can use the C1V1=C2V2 equation.

C1 = starting concentration.

C2 = final concentration.

V is for volume.

So let's say you have 500 mL of 1 M HCl and you only want 50 mL of 0.1 M HCl.

1 M x V1 = 0.1 M x 50 mL

V1 = 5 mL

So you take 5 mL of the original (stock) solution and dilute it up to 50 mL. Then you have 50 mL of 0.1 M HCl.

Explanation:

4 0

3 years ago

Which salt is formed by the reaction between hydrochloric acid and sodium hydroxide, write with chemical equation. .

stepan [7]

Answer:

salt sodium chloride (NaCl)

8 0

3 years ago

How many orbitals are in the 5th energy level?

Alexandra [31]

Answer:7

Explanation:

8 0

3 years ago

Read 2 more answers

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