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kozerog [31]
3 years ago
15

How many protons, neutrons, and electrons does sodiums neutral atom have? Plz help.

Chemistry
1 answer:
rusak2 [61]3 years ago
3 0

Answer:

So for your question, the Periodic Table tells us that sodium has an Atomic Number of 11, so there are 11 protons and 11 electrons. The Periodic Table tells us that sodium has an Atomic Mass of ≈23. So there are 23 - 11 = 12 neutrons.

Explanation:

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How many molecules are in 2.38g of SO2
Murrr4er [49]

2.63 x 10^22 molecules of SO2.

To find this, start with what you know.

2.38g of SO2.

You need to first convert this into Moles since you cannot directly convert grams into molecules. In order to convert grams to moles, you need to find the molecular mass of SO2 - 64.066.

This is because Sulfur has the mass of 32.066 and Oxygen has a mass of 16, but since there are two Oxygen atoms, it's going to be 32.

Your equation should currently appear as so:

2.38g of SO2 = 1 Mole of SO2 / 64.066

Now, you need to convert this to molecules.

Whenever you are searching for molecules without a given amount, you always use Avogadro's number: 6.02 x 10^23

Now, your equation should appear as so:

2.38g SO2 = 1 Mole of SO2 / 64.066 next to a new fraction which is 6.02 x 10^23 / 1 Mole of SO2

Now, multiply across (2.38 x 1 x 6.02 x 10^23). When using Avogadro's number, don't forget to place parenthesis around it.

Then, divide that number by the bottom: 64.066.

Thus, your final answer is that there is 2.63 x 10^22 molecules of SO2.

Don't forget your units!

Hope this helps!

5 0
3 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

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Answer:

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