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3 years ago

According to Bohr atomic model

Chemistry

1 answer:

nata0808 [166]3 years ago

7 0

Answer:

A small positively charged nucleus surrounded by revolving negatively charged electrons in fixed orbits

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Answer:

the awnser is B

Explanation:

the awnser is B

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2 years ago

The shielding of electrons gives rise to an effective nuclear charge, Zeff, which explains why boron is larger than oxygen. Esti

Paul [167]

Answer:

Option B. +3 and +6

Explanation:

Zeff = z - s

where z is the atomic number, s is the number of shielding(non-valence) electrons

For Boron, electronic configuration is 1s²2s³.

z = 5, s = 2

Zeff = 5 - 2 = +3

For Oxygen, electronic configuration is 1s²2s²2p⁴

z = 8, s = 2

Zeff = 8 - 2 = +6

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3 years ago

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Answer:

Trials A and B

Explanation:

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3 years ago

CH4 with pressure 1 atm and volume 10 liter at 27°C is passed into a reactor with 20% excess oxygen, how many moles of oxygen is

BaLLatris [955]

Answer : The moles of $O_2$ left in the products are 0.16 moles.

Explanation :

First we have to calculate the moles of $CH_4$ .

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = 1 atm

V = volume of gas = 10 L

T = temperature of gas = $27^oC=273+27=300K$

n = number of moles of gas = ?

R = gas constant = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

$(1atm)\times (10L)=n\times (0.0821L.atm/mol.K)\times (300K)$

$n=0.406mole$

Now we have to calculate the moles of $O_2$ .

The balanced chemical reaction will be:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

From the balanced reaction we conclude that,

As, 1 mole of $CH_4$ react with 2 moles of $O_2$

So, 0.406 mole of $CH_4$ react with $2\times 0.406=0.812$ moles of $O_2$

Now we have to calculate the excess moles of $O_2$ .

$O_2$ is 20 % excess. That means,

Excess moles of $O_2$ = $\frac{(100 + 20)}{100}$ × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × Required moles of $O_2$

Excess moles of $O_2$ = 1.2 × 0.812 = 0.97 mole

Now we have to calculate the moles of $O_2$ left in the products.

Moles of $O_2$ left in the products = Excess moles of $O_2$ - Required moles of $O_2$

Moles of $O_2$ left in the products = 0.97 - 0.812 = 0.16 mole

Therefore, the moles of $O_2$ left in the products are 0.16 moles.

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