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Serhud [2]
4 years ago
15

Benzene is 92.3% carbon and 7.7% hydrogen. In anexperiment,

Chemistry
1 answer:
Marysya12 [62]4 years ago
7 0

Answer:

The molecular formula of benzene = C_{6}H_{6}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

% of C = 92.3

Molar mass of C = 12.0107 g/mol

% moles of C = \frac{92.3}{12.0107} = 7.6848

% of H = 7.7

Molar mass of H = 1.00784 g/mol

% moles of H = \frac{7.7}{1.00784} = 7.6401

Taking the simplest ratio for C and H as:

7.6848 : 7.6401

= 1 : 1

The empirical formula is = CH

Molecular formulas is the actual number of atoms of each element in the compound while empirical formulas is the simplest or reduced ratio of the elements in the compound.

Thus,  

Molecular mass = n × Empirical mass

Where, n is any positive number from 1, 2, 3...

Mass from the Empirical formula = 12+ 1 = 13 g/mol

Molar mass = 78.0 g/mol

So,  

Molecular mass = n × Empirical mass

78.0 = n × 13

⇒ n ≅ 6

<u>The molecular formula of benzene = C_{6}H_{6}</u>

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<u>Answer:</u>    A. constant

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<u>Explanation:</u>

Given that when the ball is released,<em>there is no air resistance acting on ball. So the ball is freely falling. </em>

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8 0
3 years ago
1.
Serhud [2]

Explanation:

Given that,

The frequency of electromagnetic spectrum is 2.73\times 10^{16}\ Hz

(A) Let the wavelength of this radiation is \lambda. We know that,

c=f\lambda\\\\\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{2.73\times 10^{16}}\\\\\lambda=1.09\times 10^{-8}\ m

So, the wavelength of this radiation is 1.09\times 10^{-8}\ m.

(B) Let E is the energy associated with this radiation. Energy of an electromagnetic radiation is given by :

E=hf

h is Planck's constant

E=6.63\times 10^{-34}\times 2.73\times 10^{16}\\\\E=1.8\times 10^{-17}\ J

1 kcal = 4184 J

It means,

1.8\times 10^{-17}\ J=\dfrac{1}{4184}\times 1.8\times 10^{-17}\\\\=4.3\times 10^{-21}\ \text{kcal}

Hence, this is the required solution.

3 0
4 years ago
Percent occurrence and isotope masses for oxygen: 99.759 % at 15.99491 amu, 0.037% at 16.99913 amu and 0.204% is at what amu val
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Answer:

15.99937 is the value of this question answer

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3 years ago
The ionization potential of Be atom is more than expected because it has -
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Answer:

(b) fully filled valence s orbitals

Explanation:

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2s2 is fully filled

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3 years ago
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expeople1 [14]

Answer : The ratio of p_{NO_2} to p_{NO} is, 6.87\times 10^5

Solution :  Given,

K_p=1.5\times 10^6

p_{O_2} = 0.21 atm

The given equilibrium reaction is,

NO(g)+\frac{1}{2}O_2\rightleftharpoons NO_2(g)

The expression of K_p will be,

K_p=\frac{(p_{NO_2})}{(p_{NO})\times (p_{O_2})^{\frac{1}{2}}}

Now put all the given values in this expression, we get:

1.5\times 10^6=\frac{(p_{NO_2})}{(p_{NO})\times (0.21)^{\frac{1}{2}}}

\frac{(p_{NO_2})}{(p_{NO})}=(1.5\times 10^6)\times (0.21)^{\frac{1}{2}}

\frac{(p_{NO_2})}{(p_{NO})}=6.87\times 10^5

Therefore, the ratio of p_{NO_2} to p_{NO} is, 6.87\times 10^5

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