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NNADVOKAT [17]
3 years ago
15

What can you conclude from the figure shown?

Chemistry
1 answer:
arlik [135]3 years ago
6 0

Answer:

A. Hund's rule has been violated

Explanation:

There must be one electron with the same spin in each orbital of the same energy before you can put two in the same orbital. In the photo, the 2s sublevel is completely filled before the 1s sublevel (if anything the 1s should have an up spin and down spin, with the 2s having one up spin).

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How many joules of heat are required to raise the temperature of 174g of gold from 22°C to 85°C? The specific heat of gold is 0.
Slav-nsk [51]
Remembering the equation Q=MCdeltaT where
q=is the amount of heat energy
M=mass
C=specific heat
deltaT= change in temperature

Therefore, using the equation we can substitute values and solve for q.
Q= (15 grams) (0.129 J/(gx°C))(85-22)
Q=(15) ((0.129 J/(gx°C)) (63)
Q=121.9 Joules

The energy needed to raise the temperature of 15 grams of gold from 22 degrees Celsius to 85 degrees Celsius is then 121.9 Joules or 122 Joules (if rounded up).

7 0
3 years ago
This chemical equation is not balanced. Which element is not conserved?
ch4aika [34]

Answer: CI

Explanation:

4 0
3 years ago
Read 2 more answers
2) Show the calculation of Kc for the following reaction if an initial reaction mixture of 0.800 mole of CO and 2.40 mole of H2
nadezda [96]

Answer:

Kc = 3.90

Explanation:

CO reacts with H_2 to form CH_4 and H_2O. balanced reaction is:

CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

No. of moles of CO = 0.800 mol

No. of moles of H_2 = 2.40 mol

Volume = 8.00 L

Concentration = \frac{Moles}{Volume\ in\ L}

Concentration of CO = \frac{0.800}{8.00} = 0.100\ mol/L

Concentration of H_2 = \frac{2.40}{8.00} = 0.300\ mol/L

                 CO(g) + 3H_2 (g) \leftrightharpoons CH_4(g)  +  H_2O(g)

Initial            0.100      0.300             0   0

equi.            0.100 -x    0.300 - 3x     x    x

It is given that,

at equilibrium H_2O (x) = 0.309/8.00 = 0.0386 M

So, at equilibrium CO = 0.100 - 0.0386 = 0.0614 M

At equilibrium H_2 = 0.300 - 0.0386 × 3 = 0.184 M

At equilibrium CH_4 = 0.0386 M

Kc=\frac{[H_2O][CH_4]}{[CO][H_2]^3}

Kc=\frac{0.0386 \times 0.0386}{(0.184)^3 \times 0.0614} =3.90

8 0
3 years ago
Two solids of identical mass, A and B, are analyzed using identical calorimeters. Each calorimeter contains the same amount of w
DIA [1.3K]

Answer:

Specific heat of solid A is greater than specific heat of solid B.

Explanation:

In the calorimeter, as the temperature is increasing, the vibrational kinetic energy will increase and this means that additional amount of energy will be needed to increase the temperature by the same value. Therefore, we can conclude that specific heat increases as temperature increases.

Now, we are told that the final temperature of solid A's calorimeter is higher than that of B.

This means from our definition earlier, Solid A will have a higher specific heat that solid B.

5 0
2 years ago
The Sun heats land and water, how is this related to wind and ocean currents?
Yuki888 [10]

Answer:

"The sun warms up parts of the oceans. Warm waters rise just like warm air rises. So, as the warmer ocean waters begin to rise in a particular area, the cooler ocean waters from a different area will move in to replace the warmer ocean waters, and this creates our ocean currents."

Explanation:

Hope this is helpful :)

8 0
2 years ago
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