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2 years ago

List the energy levels for the orbital configuration of vanadium (v) atomic number 23.

Chemistry

1 answer:

qwelly [4]2 years ago

8 0

Answer:

n=1,2,3,4

Explanation:

The electronic configuration of vandanium is

1s2 2s2 2p6 3s2 3p6 3d3 4s2

There are two electrons in n=1, eight electrons in n=2, eleven electrons in n=3 and two electrons in n=4

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To 100.0 g water at 25.00 ºc in a well-insulated container is added a block of aluminum initially at 100.0 ºc. the temperature o

11111nata11111 [884]

When the amount of heat gained = the amount of heat loss

so, M*C*ΔTloses = M*C* ΔT gained

when here the water is gained heat as the Ti = 25°C and Tf= 28°C so it gains more heat.

∴( M * C * ΔT )W = (M*C*ΔT) Al

when Mw is the mass of water = 100 g

and C the specific heat capacity of water = 4.18

and ΔT the change in temperature for water= 28-25 = 3 ° C

and ΔT the change in temperature for Al = 100-28= 72°C

and M Al is the mass of Al block

C is the specific heat capacity of the block = 0.9

so by substitution:

100 g * 4.18*3 = M Al * 0.9*72

∴ the mass of Al block is = 100 g *4.18 / 0.9*72

= 19.35 g

4 0

2 years ago

The volume of a gas at 7.00°c is 49.0 ml. if the volume increases to 74.0 ml and the pressure is constant, what will the tempera

marshall27 [118]

Using charles law
v1/t1=v2/t2
v1=49ml
v2=74
t1=7+273=280k
t2=?
49/280=74/t2
0.175=74/t2 cross multiply
0.175t2=74
t2=74/0.175
t2=422k or 149celcius

8 0

1 year ago

A beach has a supply of sand grains composed of calcite, ferromagnesian silicate minerals, and non-ferromagnesian silicate miner

bezimeni [28]

Ferromagnesian silicate minerals (i looked it up)

4 0

1 year ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Explanation:

For 1: At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

For 2:

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$

So, $1.7\times 10^{-4}mol$ of methyl amine will produce = $\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+$

To calculate the $pK_b$ of base, we use the equation:

$pK_b=-\log(K_b)$

where,

$K_b$ = base dissociation constant = $3.9\times 10^{-10}$

Putting values in above equation, we get:

$pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41$

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})$

We are given:

$pK_b=9.41$

$[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

$[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

Putting values in above equation, we get:

$pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-9.41=4.59$

Hence, the pH of the solution is 4.59

4 0

2 years ago

In this problem, you will use Lenz's law to explore what happens when an electromagnet is activated a short distance from a wire

Vsevolod [243]

Answer:

The correct answer is Option A (There is no magnetic flux through the wire loop.)

Explanation:

Magnetic flux measures the entire magnetic field that passes through the wire loop.

The right hand rule can be demonstrated on how magnetic flux is generated through the moving current in the wire loop. The magnetic flux through the wire loop will decrease as it moves upward through the magnetic field region.

If the direction of the vector area of the wire loop is to the right, and the switch is closed, it will push the magnetic flux to the right which will now be increased due to an equal increase in the current in the wire loop. But, when the switch is open, this will halt the movement of current through the wire loop thus affecting the generation of magnetic field. This would make the magnetic flux to be zero.

4 0

2 years ago