1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Tju [1.3M]
3 years ago
14

How many valence electrons does Ne have

Chemistry
2 answers:
ollegr [7]3 years ago
6 0

Answer:

eight valence electrons

lidiya [134]3 years ago
5 0

Answer:

it has 8 valence electrons

You might be interested in
Write the rates for the following reactions in terms of the disappearance of reactants and appearance of products: (a) 302 .....
Oduvanchick [21]

Answer:

Explanation:

\mathbf{From \  the  \ information \  given:} \\ \\ \mathbf{The \  rates \  of  \ the \ f ollowing \  reactions \  can \  be \  expressed  \ as \  follows:}

(a)

\mathbf{3O_2 \to 2O_3} \\ \\ \\ \mathbf{-\dfrac{1}{3}\dfrac{d[O_2]}{dt}=\dfrac{2}{3} \dfrac{d[O_3]}{dt}}

(b)

\mathbf{C_2H_6 \to C_2H_4 + H_2}  \\ \\ \\ \mathbf{  -\dfrac{d[C_2H_6]}{dt}= \dfrac{d[C_2H_4]}{dt}=\dfrac{d[H_2]}{dt}}

(c)

\mathbf{ClO^-+Br^- \to BrO^-+Cl^-} \\ \\ \\ \mathbf{ -\dfrac{d[ClO^-]}{dt}= -\dfrac{d[Br^-]}{dt} =  \dfrac{d[BrO^-]}{dt} = \dfrac{d[Cl^-]}{dt}   }

(d)

\mathbf{(CH_3)_3 CCl+H_2O \to (CH_3)_3COH + H^+ + Cl^-} \\ \\ \\  \mathbf{- \dfrac{d[(CH_3)_3CCl}{dt}= - \dfrac{d[H_2O]}{dt}= \dfrac{d[CH_3)_3COH}{dt}= \dfrac{d[H^+]}{dt}= \dfrac{d[Cl^-]}{dt}}

(e)

\mathbf{2AsH_3 \to 2As + 3H_2} \\ \\ \\  \mathbf{-\dfrac{1}{2}\dfrac{d[AsH_3]}{dt}=\dfrac{1}{2}\dfrac{d[As]}{dt}=\dfrac{1}{3}\dfrac{d[H_2]}{dt}}

5 0
3 years ago
The chemical equation is Mg(s) + O2(g) → MgO(s)
vredina [299]

Answer:

The magnesium will burn until consumed entirely. There is much more oxygen available in the atmosphere than needed to consume the magnesium. Thus the magnesium is the limiting reactant because it determines the amount of product formed.

Explanation:

Mg produces less amount of MgO than O2; therefore Mg is the limiting reagent. O2 produces more amount of MgO than Mg; therefore O2 is the excess reagent.

4 0
3 years ago
Water's ability to dissolve a wide variety of molecules is important, but more important is the hydrophobic effect, which drives
Sliva [168]

Answer:

d. Hydrophobic molecules are attracted to each other.

Explanation:

The term “hydrophobic effect” is associated with the spontaneous tendency of macromolecules, such as proteins, to prefer a conformation in an aqueous medium, with hydrophobic groups facing the interior of the mac romolecule, favoring attractive intramolecular interactions, and hydrophilic groups exposed on the surface, for maximize interactions with water molecules in the medium. This is because the hydrophobic molecules are attracted to each other, allowing them to turn inward.

8 0
3 years ago
Thermochemistry
max2010maxim [7]
THe Answer is 105 J

hope this helps and please brainliest, I was first :)
4 0
3 years ago
You dissolve 8.65 grams of lead(l) nitrate in water and then you add 2 50 grams of aluminum. This reaction occurs 2AI(S)+ 3Pb(NO
olga55 [171]

<u>Answer:</u> The theoretical yield of solid lead comes out to be 5.408 grams.

<u>Explanation:</u>

To calculate the moles, we use the following equation:  

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}  

  • <u>Moles of Lead nitrate:</u>

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

\text{Number of moles}=\frac{8.65g}{331.2g/mol}=0.0261moles

  • <u>Moles of Aluminium:</u>

Given mass of aluminium = 2.5 grams

Molar mass of aluminium = 27 g/mol

Putting values in above equation, we get:

\text{Number of moles}=\frac{2.5g}{27g/mol}=0.0925moles

For the given chemical reaction, the equation follows:

2AI(s)+3Pb(NO_3)_2(aq.)\rightarrow 3Pb(s)+2AI(NO3)_3(aq.

By Stoichiometry:

3 moles of lead nitrate reacts with 2 moles of aluminium

So, 0.0261 moles of lead nitrate are produced by = \frac{2}{3}\times 0.0261=0.0174moles of aluminium.

As, the required amount of aluminium is less than the given amount. Hence, it is considered as the excess reagent.

Lead nitrate is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

3 moles of lead nitrate are produces 3 moles of lead metal.

So, 0.0261 moles of lead nitrate will produce = \frac{3}{3}\times 0.0261=0.0261moles of lead metal.

  • Now, to calculate the grams or theoretical yield of lead metal, we put in the mole's equation, we get:

Molar mass of lead = 207.2 g/mol

Putting values in above equation, we get:

0.0261mol=\frac{\text{Given mass}}{207.2g/mol}

Mass of lead = 5.408 grams

Hence, the theoretical yield of solid lead comes out to be 5.408 grams.

8 0
3 years ago
Other questions:
  • If 0.300 moles of Cl2 is mixed with 0.600 moles of F2 in a 5.00 L rigid reaction vessel, what is the final pressure in the vesse
    13·1 answer
  • What type of claim must be supported by scientifically valid evidence for it to be approved for use on a food label? A. Health c
    13·2 answers
  • What are close-toed shoes least likely to provide protection against?
    6·2 answers
  • More than 70% of all legislation passed in Washington, DC addresses issues related to science; most of these issues pertain chem
    9·1 answer
  • Determine the molarity for the fixer Sodium carbonate decahydrate (I dissolved 127.62 grams of sodium carbonate for every 4 lite
    13·1 answer
  • If the volume of a cylinder is 84,300 cubic centimeters then how many millimeters is this equal to
    7·1 answer
  • What type of pressure would you expect to find at the equator?
    7·1 answer
  • What is true about most of the wavelengths found in the electromagnetic spectrum?
    12·1 answer
  • Explain, in full sentences, in terms of heat transfer (what type of heat transfer /how do you know?) AND specific heat (why copp
    7·1 answer
  • BaO2 is a peroxide. Why?
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!