1. How does the figure below illustrate Hund's rule?

Chemistry

1 answer:

Elan Coil [88]3 years ago

6 0

Answer:

it indicates that the lower orbitals are filled first and they have the maximum number of electrons compared to the higher orbitals

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What is the mass of the question?

musickatia [10]

Answer: I do not know

Explanation:

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4 years ago

Consider the following unbalanced equation. N2+ O2=N2O. When the equation is completely balanced using smallest whole numbers, t

77julia77 [94]

Answer:

Explanation:

Since you are balancing the equation you need to have equal amounts of each element on each side. Right now there is 2n and 2o on one side, and on the other 2n and 1o. Although the number of n's on each side is equal the number of o's are not. First I balance the number of oxygen's on each below, but doing that made the number of nitrogen's unbalanced. So then I balanced the nitrogens in step 2.

__N2+ __O2 = __N2O

1. __N2+ __O2 = 2 N20

2. 2 N2 + 1 O2 = 2 N2O

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3 years ago

A 30-0 g sample of water at 280 K is mixed with 50.0 g of water at 330 K. How would you calculate the final temperature of the m

aliya0001 [1]

Answer:

311.25k

Explanation:

The question assumes heat is not lost to the surroundings, therefore

heat emitted from hotter sample ( $q_{\ lost}$ )= heat absorbed by the less hotter sample( $q_{\ gain}$ )

The relationship between heat (q), mass (m) and temperature (t) is $q = mc\Delta t$

where c is specific heat capacity, $\Delta t$ temperature change.

$\Delta t$ = $t_{\ final} - t_{\ initial}$

equating both heat emitted and absorb

$-q_{\ lost} = q_{\ gain}$

$-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})$

where the values with subset 1 are the values of the hotter sample of water and the values with subset 2 are the values of the less hot sample of water.

C will cancel out since both are water and they have the same specific heat capacity.

so we have

$-m_{1}(t_{\ final} - t_{\ 1initial})=m_{2}(t_{\ final} - t_{\ 2initial})$

where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?

-50 * ( $t_{final}$ - 330) = 30 * ( $t_{final}$ - 280)