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3 years ago

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Chemistry

1 answer:

Galina-37 [17]3 years ago

3 0

Answer:

Explanation:

defiantly

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What is the electron configuration of the chlorine atom after it gains an electron?

MatroZZZ [7]

Answer:

Check this question from quizlet :)

* do not type those number only QUESTION and then its will show you the answer with a number from quizlet *

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3 years ago

Please help I’ll mark brainliest. Thank you :)

Jobisdone [24]

Answer:

a. There are only 5 electrons in 2p. 2p should be filled with 6 electrons prior to 3s and 3p

b. There are 8 electrons in 2p. 2p should be filled only with 6 electrons (not 8)

Explanation:

The typical electron configuration: 1s2 2s2 2p6 3s2 3p6

a. There are only 5 electrons in 2p

=> should be 1s2 2s2 2p6 3s2 3p2

b. There are 8 electrons in 2p

=> should be 1s2 2s2 2p6 3s2 3p1

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2 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$

Therefore, we plug in the data to obtain:

$V=\frac{3.7mol*0.08206\frac{atm*L}{mol*K}*273.15K}{1atm}\\\\V=82.9L$

Best regards!

6 0

3 years ago

72.0 grams of water how many miles of sodium with react with it?

Flura [38]

Answer:

$\large \boxed{\text{8.00 mol}}$

Explanation:

We will need a balanced chemical equation with masses, moles, and molar masses.

1. Gather all the information in one place:

Mᵣ: 18.02

2Na + H₂O ⟶ 2NaOH + H₂

m/g: 72.0

2. Moles of H₂O

$\text{Moles of H$_{2}$O} = \text{72.0 g H$_{2}$O} \times \dfrac{\text{1 mol H$_{2}$O}}{\text{18.02 g H$_{2}$O}} = \text{3.996 mol H$_{2}$O}$

3. Moles of Na

The molar ratio is 2 mol Na/1 mol H₂O.

$\text{Moles of Na} = \text{3.996 mol H$_{2}$O} \times \dfrac{\text{2 mol Na}}{\text{1 mol H$_{2}$O}} = \textbf{8.00 mol Na}\\\\\text{The water will react with $\large \boxed{\textbf{ 8.00 mol}}$ of Na}$

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3 years ago

3. Which is the minimum amount of energy required to start a chemical reaction?

jekas [21]

B. Chemical energy to start a chemical reaction

3 0

3 years ago

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