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3 years ago

What is the electron structure of a sodium atom?

Chemistry

2 answers:

Alla [95]3 years ago

7 0

1s22s22p63s1 is the electronic configuration of sodium.

labwork [276]3 years ago

6 0

Ans: A)

The electron configuration of an element can be represented based on Aufbau principle according to which the electrons fill up from the lower to higher energy levels.

s orbitals can accommodate a maximum of 2 electrons

p orbitals = 6 e-

d orbitals = 10 e-

f orbitals = 14 e-

The atomic number for sodium (Na) = 11

The 11 electrons in Na can be arranged as follows:

Na[11] = 1s²2s²2p⁶3s¹

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For a constant waves speed,wavelength,of the wave increases with a(n)______

pochemuha

Answer:

higher frequency

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3 years ago

Acid-catalyzed dehydration of secondary and tertiary alcohols proceeds through an E1 mechanism. The first step is the protonatio

Over [174]

Answer:

Most substituted alkene is produced as a major product

Explanation:

Dehydration of 3-methyl-2-butanol proceeds through E1 mechanism to form alkenes.
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2 years ago

N₂O(g) + 3 H₂(g) N₂H4(1) + H₂O(1) AH = -317 kJ/mol

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Answer:

Explanation:

Recall that Δ<em>H</em> is the sum of the heats of formation of the products minus the heat of formation of the reactants multiplied by their respective coefficients. That is:

$\displaystyle \Delta H^\circ_{rxn} = \sum \Delta H^\circ_{f} \left(\text{Products}\right) - \sum \Delta H^\circ_{f} \left(\text{Reactants}\right)$

Therefore, from the chemical equation, we have that:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) = \left[\Delta H^\circ_f \text{ N$_2$H$_4$} + \Delta H^\circ_f \text{ H$_2$O} \right] -\left[3 \Delta H^\circ_f \text{ H$_2$}+\Delta H^\circ_f \text{ N$_2$O}\right] \end{aligned}$

Remember that the heat of formation of pure elements (e.g. H₂) are zero. Substitute in known values and solve for hydrazine:

$\displaystyle \begin{aligned} (-317\text{ kJ/mol}) & = \left[ \Delta H^\circ _f \text{ N$_2$H$_4$} + (-285.8\text{ kJ/mol})\right] -\left[ 3(0) + (82.1\text{ kJ/mol})\right] \\ \\ \Delta H^\circ _f \text{ N$_2$H$_4$} & = (-317 + 285.8 + 82.1)\text{ kJ/mol} \\ \\ & = 50.9\text{ kJ/mol} \end{aligned}$

In conclusion, our answer is A.

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2 years ago

Need help with these.

kakasveta [241]

I got you bro by the power of my heart and the felling of agony my condolences go to the Siskin of Thebes

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3 years ago

How much energy is needed to completely boil a 5.05g sample of water?

Keith_Richards [23]

Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

<h3>Why does it take this much energy to boil the water?</h3>

We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.

Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.

To learn more about the behavior of water visit:

brainly.com/question/1416592?referrer=searchResults

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2 years ago