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3 years ago

List first thirty elements with their valences

Chemistry

1 answer:

lbvjy [14]3 years ago

5 0

Element Atomic Number Valency

Valency of Hydrogen 1 1

Valency of Helium 2 0

Valency of Lithium 3 1

Valency of Beryllium 4 2

Valency of Boron 5 3

Valency of Carbon 6 4

Valency of Nitrogen 7 3

Valency of Oxygen 8 2

Valency of Fluorine 9 1

Valency of Neon 10 0

Valency of Sodium (Na) 11 1

Valency of Magnesium (Mg) 12 2

Valency of Aluminium 13 3

Valency of Silicon 14 4

Valency of Phosphorus 15 3

Valency of Sulphur 16 2

Valency of Chlorine 17 1

Valency of Argon 18 0

Valency of Potassium (K) 19 1

Valency of Calcium 20 2

Valency of Scandium 21 3

Valency of Titanium 22 4

Valency of Vanadium 23 5,4

Valency of Chromium 24 2

Valency of Manganese 25 7, 4, 2

Valency of Iron (Fe) 26 2, 3

Valency of Cobalt 27 3, 2

Valency of Nickel 28 2

Valency of Copper (Cu) 29 2, 1

Valency of Zinc 30 2

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Answer: The longest wavelength of light is 656.5 nm

Explanation:

For the longest wavelength, the transition should be from n to n+1, where: n = lower energy level

To calculate the wavelength of light, we use Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )$

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant = $1.096776\times 10^7m^{-1}$

$n_f$ = Higher energy level = $n_i+1=(2+1)=3$

$n_i$ = Lower energy level = 2 (Balmer series)

Putting the values in above equation, we get:

$\frac{1}{\lambda }=1.096776\times 10^7m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\\\\\lambda =\frac{1}{1.5233\times 10^6m^{-1}}=6.565\times 10^{-7}m$

Converting this into nanometers, we use the conversion factor:

$1m=10^9nm$

So, $6.565\times 10^{-7}m\times (\frac{10^9nm}{1m})=656.5nm$

Hence, the longest wavelength of light is 656.5 nm

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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

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Answer: The theoretical yield of solid lead comes out to be 5.408 grams.

Explanation:

To calculate the moles, we use the following equation:

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

Moles of Lead nitrate:

Given mass of lead nitrate = 8.65 grams

Molar mass of lead nitrate = 331.2 g/mol

Putting values in above equation, we get:

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

Moles of Aluminium:

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.

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List first thirty elements with their valences​

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