How should i read chemistry ?

Which types of electron orbitals will have higher energy than a 4d orbital?

alexdok [17]

Answer:

D) 4f

Explanation:

To determine which electron orbital that will have higher energy than a 4d orbital, we write the electron configuration starting with s-orbital.

1s

2s 2p

3s 3p 3d 3f

4s 4p 4d 4f

5s 5p 5d 5f

6s 6p 6d 6f

7s 7p 7d 7f

In ascending order, 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 3f, <u>4d</u><em>,</em> 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

From the electronic configuration formula above, the electron orbitals that have higher energy than a 4d orbital are 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 6f, 7d, 7f.

Therefore, 4f is the correct answer.

8 0

3 years ago

The states of matter involved in a milkshake

Anna [14]

Answer:

it is a liquid that is made from the combination of a liquid and a solid.

Explanation:

When these two states of matter are ground and mixed together, they form one state that is liquid

6 0

2 years ago

Read 2 more answers

Given that kb for c6h5nh2 is 1.7 × 10-9 at 25 °c, what is the value of ka for c6h5nh3 at 25 °c?

lina2011 [118]

Answer: $k_a$ for $C_6H_5NH_3^+$ at 25°C is $0.588\times 10^{-4}$

Explanation: We are given $k_b$ of $C_6H_5NH_2$ at 25°C which is $1.7\times 10^{-9}$

To calculate the $k_a$ of $C_6H_5NH_3^+$ , we use the formula:

$k_w=k_a\times k_b$

$k_w\text{ at }25^o=1\times 10^{-14}$

Putting values in above equation, we get:

$1\times 10^{-14}=k_a\times (1.7\times 10^{-9})\\k_a=0.588\times 10^{-5}$

7 0

3 years ago

Lead(II) nitrate and ammonium iodide react to form lead(II) iodide and ammonium nitrate according to the reaction Pb ( NO 3 ) 2

nikklg [1K]

Answer : The volume of $NH_4I$ solution required is, 2.93 L

The number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

Explanation :

First we have to calculate the initial moles of $Pb(NO_3)_2$ .

$\text{Moles of }Pb(NO_3)_2=\text{Concentration of }Pb(NO_3)_2\times \text{Volume of solution}$

$\text{Moles of }Pb(NO_3)_2=0.700M\times 0.945L=0.662mol$

Now we have to calculate the moles of $NH_4I$

The balanced chemical reaction is:

$Pb(NO_3)_2(aq)+2NH_4I(aq)\rightarrow PbI_2(s)+2NH_4NO_3(aq)$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react with 2 moles of $NH_4I$

So, 0.662 mole of $Pb(NO_3)_2$ react with $0.662\times 2=1.32$ moles of $NH_4I$

Now we have to calculate the volume of $NH_4I$

$\text{Volume of }NH_4I=\frac{\text{Moles of }NH_4I}{\text{Concentration of }NH_4I}$

$\text{Volume of }NH_4I=\frac{1.32mol}{0.450mol/L}=2.93L$

Now we have to calculate the moles of $PbI_2$

From the balanced chemical reaction we conclude that,

As, 1 mole of $Pb(NO_3)_2$ react to give 1 moles of $PbI_2$

So, 0.662 mole of $Pb(NO_3)_2$ react to give 0.662 moles of $PbI_2$

Thus, the number of moles of $PbI_2$ formed from the reaction is, 0.662 moles.

7 0

3 years ago

Balance the following chemical equations

nexus9112 [7]

Explanation:

2Na + I₂ → 2NaI
2KClO₃ → 2KCl + 3O₂
K₃PO₄ + 3HCl → 3KCl + H₃PO₄

6 0

3 years ago

How should i read chemistry ? ​

How should i read chemistry ?