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3 years ago

What’s the distance from the nucleus of an atom to its theoretical outer edge of electron orbits

Chemistry

1 answer:

ollegr [7]3 years ago

4 0

Answer:

The range of atoms = (30-300 pm) depending upon the element

Explanation:

The Atomic radii of the atom is the distance from the center of the circle to the outermost orbital.

The center of the circle is the nucleus and the radii is the outermost boundary.

The actual size of the atom is decided on the basis of the Zeff . Also known as effective nuclear charge.

Zeff: It is the net positive charge felt by the outermost electron by the nucleus.

The value of Zeff depends upon the shielding constant. More the shielding less will be the Zeff . Hence the size of the atom increases.

Due to shielding the outermost electrons feel less pull of nucleus.

The greater the Zeff , the smaller the radius of the atom.

The formula used to calculate the atomic mass is :

$r_{n}=\frac{52.9n^{2}}{Z}$ pm

Here "pm"= picometers

$1 pm = 10^{-12}m$

The size of the smallest atom H-atom = 120 pm

The range of atoms = (30-300 pm)

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When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin

loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=T_f^0-T_f=8.2^0C$ = Depression in freezing point

$K_f$ = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality = $\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9$

$8.2^0C=1\times K_f\times 1.9$

$K_f=4.32^0C/m$

Now Depression in freezing point for sodium chloride is given by:

$\Delta T_f=i\times K_f\times m$

$\Delta T_f=20.0^0C$ = Depression in freezing point

$K_f$ = freezing point constant

m= molality = $\frac{136g\times 1000}{950g\times 58.5g/mol}=2.45$

$20.0^0C=i\times 4.32^0C\times 2.45$

$i=1.9$

Thus vant hoff factor for sodium chloride in X is 1.9

3 0

3 years ago

Is pyrophoric a chemical or physical property

Neporo4naja [7]

Answer:

Pyrophoricity is a property of metals and oxides of lower oxidation states, including radioactive ones, in which they spontaneously ignite during or after stabilization.

5 0

3 years ago

N writing a chemical equation that produces hydrogen gas, the correct representation of hydrogen gas is

andrezito [222]

Need more information

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3 years ago

Which element is oxidized in the reaction below? fe(co)5 (l) + 2hi (g) fe(co)4i2 (s) + co (g) + h2 (g)?

DanielleElmas [232]

Oxidation state of I is (-1) and for CO it is zero. Let's assume that the oxidation state of Fe in Fe(CO)₄I₂ (s) is x. For whole compound, the charge is zero.

Sum of oxidation numbers in all elements = Charge of the compound.

Here we have 1Fe , 4CO and 2I
hence we can find the oxidation state as;
x + 4*0 + 2*(-1) = 0
x + 0 - 2 = 0
x = +2
Hence the oxidation state of Fe in product Fe(CO)₄I₂ (s) is +2.

Same as we can find the oxidation state (y) of Fe in Fe(CO)₅(s).
y + 5*0 = 0
y = 0

Since oxidation state of Fe increased from 0 to +2, the oxidized element is Fe in the given reaction.

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3 years ago

What variable represents specific heat in the equation Q = mcAT?

Elan Coil [88]

Answer: A. The variable c

============================================

Explanation:

Q = heat transferred

m = mass

c = specific heat

$\Delta T$ = delta T = change in temperature

4 0

3 years ago