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Nookie1986 [14]

4 years ago

15

Why do electrons enter the 4s orbital before entering the 3d orbital

1 answer:

-BARSIC- [3]4 years ago

5 0

According to the Aufbau principle, the 4s sublevel<span> is filled </span>before the 3d sublevel <span>because the </span>4s<span> is lower in energy. ... This is why when </span>electrons<span> are lost from the </span>orbitals<span> of the transition metals, they are lost from the </span>4s<span> first because it is higher in energy</span>

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A diver exhales a bubble with a volume of 250 mL at a pressure of 2.4 atm and a temperature of 15 °C. What is the volume of the

alexdok [17]

Answer : The volume of the bubble is, 625 mL

Explanation :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 2.4 atm

$P_2$ = final pressure of gas = 1.0 atm

$V_1$ = initial volume of gas = 250 mL

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas = $15^oC=273+15=288K$

$T_2$ = final temperature of gas = $27^oC=273+27=300K$

Now put all the given values in the above equation, we get:

$\frac{2.4atm\times 250mL}{288K}=\frac{1.0atm\times V_2}{300K}$

$V_2=625mL$

Therefore, the volume of the bubble is, 625 mL

7 0

3 years ago

2. Calculate the density of a metal that occupies 17.75 cm and has a mass of 342.93 g. [D = m/V]

satela [25.4K]

Answer:

19.32

Explanation:

the density is given by the mass over volume

the mass In this case is 342.93 and the volume is 17.75

d=342.93g/17.75cm

=19.32g/cm

i hope this helps

3 0

3 years ago

Consider the following combustion reaction:

ZanzabumX [31]

Answer:

vxcvggccbbcCbnmhxx vgfscvjjg

6 0

3 years ago

Read 2 more answers

What was thompson working with when he discovered the cathode rays?

Viefleur [7K]

He was working with electrons

5 0

4 years ago

Read 2 more answers

A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after

choli [55]

Answer:

9.25

Explanation:

Let first find the moles of $NH_4Cl$ and $NaOH$

number of moles of $NH_4Cl$ = 0.40 mol/L × 200 × 10⁻³L

= 0.08 mole

number of moles of $NaOH$ = 0.80 mol/L × 50 × 10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

The ICE Table is shown below as follows:

$NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}$

Initial (M) 0.08 0.04 0

Change (M) - 0.04 -0.04 + 0.04

Equilibrium (M) 0.04 0 0.04

$K_a*K_b = 10^{-14} \ at \ 25^0C$

$K_a = \frac{10^{-14}}{K_b}$

$K_a = \frac{10^{-14}}{1.76*10^{-5}}$

$K_a= 5.68*10^{10}$

$pK_a = - log \ (K_a)$

$pK_a = - log \ (5.68*10^{-10})$

$pK_a = 9.25$

$pH = pKa + \ log (\frac{HB}{HA} )$ for buffer solutions

$pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} )$ since they are in the same solution

$pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )$

$pH = 9.25$

8 0

3 years ago