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Anton [14]
3 years ago
12

What role does valence electrons play in reactivity for an element?

Chemistry
1 answer:
kirza4 [7]3 years ago
4 0

Explanation:

The valence electrons within an atom is the number of electrons in its outermost shell.

These electrons are used by an atom to react with one another. They  determine the extent to which an atom is ready to combine either by losing, gaining or sharing these electrons.

  • Every atom desires to have a completely filled outermost shell.
  • Only the elements in group 8 have a complete octet.
  • The need to attain stability is driven by the number of electrons in their valence shell.
  • Therefore, some atoms are very reactive.
  • Those needing one electrons to complete their octet and also those that must lose one electron are very reactive.
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Based on the equation, how many grams of Br2 are required to react completely with 36.2 grams of AlCl3?
s2008m [1.1K]

Answer:

65.08 g.

Explanation:

  • For the reaction, the balanced equation is:

<em>2AlCl₃ + 3Br₂ → 2AlBr₃ + 3Cl₂,</em>

2.0 mole of AlCl₃ reacts with 3.0 mole of Br₂ to produce 2.0 mole of AlBr₃ and 3.0 mole of Cl₂.

  • Firstly, we need to calculate the no. of moles of 36.2 grams of AlCl₃:

<em>n = mass/molar mass</em> = (36.2 g)/(133.34 g/mol) = <em>0.2715 mol.</em>

<u><em>Using cross multiplication:</em></u>

2.0 mole of AlCl₃ reacts with → 3.0 mole of Br₂, from the stichiometry.

0.2715 mol of AlCl₃ reacts with → ??? mole of Br₂.

∴ The no. of moles of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃  = (0.2715 mol)(3.0 mole)/(2.0 mole) = 0.4072 mol.

<em>∴ The mass of Br₂ reacts completely with 0.2715 mol (36.2 g) of AlCl₃ = no. of moles of Br₂ x molar mass</em> = (0.4072 mol)(159.808 g/mol ) = <em>65.08 g.</em>

4 0
3 years ago
Over-encouraging lesser skilled players can lead to embarrassment.<br> a. True<br> b. False
Margarita [4]
My answer would be a. True
6 0
3 years ago
Read 2 more answers
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
The radius of an atom is closest in size to a
tigry1 [53]

Answer:

The size of an isolated atom can't be measured because we can't determine the location of the electrons that surround the nucleus. We can estimate the size of an atom, however, by assuming that the radius of an atom is half the distance between adjacent atoms in a solid. This technique is best suited to elements that are metals, which form solids composed of extended planes of atoms of that element. The results of these measurements are therefore often known as metallic radii.

.Explanation:

3 0
3 years ago
Ethanol melts at -114 degree C. The enthalpy of fusion
Brut [27]

Answer: The heat required is 6.88 kJ.

Explanation:

The conversions involved in this process are :

(1):ethanol(s)(-135^0C)\rightarrow ethanol(s)(-114^0C)\\\\(2):ethanol(s)(-114^0C)\rightarrow ethanol(l)(-114^0C)\\\\(3):ethanol(l)(-114^0C)\rightarrow ethanol(l)(-50^0C)

Now we have to calculate the enthalpy change.

\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]+n\times \Delta H_{vap}+[m\times c_{p,g}\times (T_{final}-T_{initial})]

where,

\Delta H = enthalpy change = ?

m = mass of ethanol = 25.0 g

c_{p,s} = specific heat of solid ethanol= 0.97 J/gK

c_{p,l} = specific heat of liquid ethanol = 2.31 J/gK

n = number of moles of ethanol = \frac{\text{Mass of ethanol}}{\text{Molar mass of ethanol}}=\frac{25.0g}{46g/mole}=0.543mole

\Delta H_{fusion} = enthalpy change for fusion = 5.02 KJ/mole = 5020 J/mole

T_{final}-T_{initial}=\Delta T = change in temperature

The value of change in temperature always same in Kelvin and degree Celsius.

Now put all the given values in the above expression, we get

\Delta H=[25.0 g\times 0.97J/gK\times (-114-(-135)K]+0.534mole\times 5020J/mole+[25.0g\times 2.31J/gK\times (-50-(-114))K]

\Delta H=6885.93J=6.88kJ     (1 KJ = 1000 J)

Therefore, the heat required is 6.88 kJ

3 0
3 years ago
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