3 years ago

Write the electron configuration for vanadium. (Use superscript and proper formatting for full credit.)

Chemistry

1 answer:

adoni [48]3 years ago

8 0

Answer:

"Ar 3d3 4s2" is the electronic configuration for Vanadium.

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Explanation:

The expression for the calculation of heat is shown below as:-

$Q=m\times C\times \Delta T$

Where,

$Q$ is the heat absorbed/released

m is the mass

C is the specific heat capacity

$\Delta T$ is the temperature change

Thus, given that:-

Mass of lead = 118.5 g

Specific heat = 0.129 J/g°C

Initial temperature = 10 °C

Final temperature = x °C

$\Delta T=(x-10)\ ^0C$

Q = 4700 J

So,

$4700=118.5\times 0.129\times (x-10)$

$118.5\times \:0.129\left(x-10\right)=4700$

$x-10=307.46083$

$x=317.46$

<u>Thus, the final temperature is:- 317.46 °C</u>

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