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3 years ago

12

What element has the following electron configuration 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d10 5p6 6s2 4f14 5d10 6p6 7s2 5f14 6

d8?

1 answer:

musickatia [10]3 years ago

8 0

Answer:

Explanation:

1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s² 4d¹⁰ 5p⁶ 6s² 4f¹⁴ 5d¹⁰ 6p⁶ 7s² 5f¹⁴ 6d⁸

The element is Darmstadtium . It is a d block element . Its symbol is Ds . Its atomic no is 110 .

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Study the image and answer the question.

konstantin123 [22]

The image represents A COMPOUND because the molecules are BONDED CHEMICALLY.
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2 years ago

Look at the picture below. Name the isotope

andrew11 [14]

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2 years ago

What is the half life of the graphed material?

daser333 [38]

Answer:

3 hours

Explanation:

To know the the correct answer to the question given above, it is important we know the definition of half-life.

The half-life of a substance is simply defined as the time taken for half the substance to decay.

Considering the diagram given above, the initial mass of the substance is 100 g.

Half of the initial mass = 100 / 2 = 50 g

Now, we shall determine the time from the graph taken to get to 50 g.

Considering the diagram given above, the time taken to get to 50 g is 3 hours.

Therefore, the half-life of the material is 3 hours.

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3 years ago

The standard enthalpy change for the following reaction is 232 kJ at 298 K. 2 H2CO(g) 2 C(s,graphite) + 2 H2(g) + O2(g) ΔH° = 23

Georgia [21]

Answer: The standard enthalpy change for this reaction is -116 kJ

Explanation:

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

$2H_2CO(g)\rightarrow 2C(s, graphite)+2H_2(g)+O_2(g)$

$\Delta H^0=232KJ$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$)C(s, graphite)+H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2CO(g)$ $\Delta H^0'=?$

According to the Hess’s law, if we reverse the reaction then the $\Delta H$ will change its sign and if we half the reaction, then the

So, the value $\Delta H^0'$ for the reaction will be:

$\Delta H^0'=\frac{(-232kJ/mole)}{2}=-116kJ$

Hence, the standard enthalpy change for this reaction is -116 kJ

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3 years ago

When kc decreases what happens to the molecules in terms of speed space and attractive force pls help

Fudgin [204]

Kc decreases as the temperature in molecules increase.

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1 year ago