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ELEN [110]
2 years ago
6

Tell why someone should visit this planet neptune, what they might be able to see or do, what they will

Chemistry
1 answer:
leonid [27]2 years ago
7 0
362782bebehhehdhduwkekennebsbdhr
You might be interested in
Question 9 (1 point)
Juliette [100K]

Noble gases

Explanation:

     Electronic configuration 1s²  2s²   2p⁶

 The element belongs to the group of the noble gases.

  • The noble gases have complete outer shell configuration of their atoms.
  • we can infer that the configuration above is for an element in the p-block because the last sub-level filled is the p-orbital.
  • The elements therefore belongs to the p-block
  • The block is from group 111A to O
  • Only the halogens and noble gases fits this picture from the option.
  • The outer most p-subshell have three orbitals requiring 6 electrons to fill them up.
  • This makes a complete and stable configuration.
  • The highest energy level of 2 is also made up of 8 electrons, an octet.
  • This is why we can conclude that they are noble gases.

Learn more:

Noble gas brainly.com/question/1781595

#learnwithBrainly

7 0
2 years ago
PLEASE HELP ME!!!
kondor19780726 [428]
<h3>Answer:</h3>

0.424 J/g °C

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right  

Equality Properties

  • Multiplication Property of Equality
  • Division Property of Equality
  • Addition Property of Equality
  • Subtraction Property of Equality<u> </u>

<u>Chemistry</u>

<u>Thermochemistry</u>

Specific Heat Formula: q = mcΔT

  • q is heat (in Joules)
  • m is mass (in grams)
  • c is specific heat (in J/g °C)
  • ΔT is change in temperature
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] m = 38.8 g

[Given] q = 181 J

[Given] ΔT = 36.0 °C - 25.0 °C = 11.0 °C

[Solve] c

<u>Step 2: Solve for Specific Heat</u>

  1. Substitute in variables [Specific Heat Formula]:                                             181 J = (38.8 g)c(11.0 °C)
  2. Multiply:                                                                                                             181 J = (426.8 g °C)c
  3. [Division Property of Equality] Isolate <em>c</em>:                                                         0.424086 J/g °C = c
  4. Rewrite:                                                                                                             c = 0.424086 J/g °C

<u>Step 3: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

0.424086 J/g °C ≈ 0.424 J/g °C

6 0
2 years ago
What is decarboxylation??<br><br>Gimme one reaction of it..​
kotegsom [21]

Answer:

Decarboxylation is a chemical reaction that removes a carboxyl group and releases carbon dioxide (CO2). Usually, decarboxylation refers to a reaction of carboxylic acids, removing a carbon atom from a carbon chain.

Explanation:

Please give me brainlist

3 0
2 years ago
What are the major species in solution when solid ammonium bromate is dissolved in water?
DaniilM [7]
The major species in solution when solid ammonium bromate is dissolved in water is shown below

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}

7 0
3 years ago
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