Answer:
Aluminum nitrate is a salt composed of aluminum and nitric acid, belonging to a group of reactive chemicals - organic nitrate and nitrite compounds. The nitrate ion is polyatomic, meaning it is composed of two or more ions that are covalently bonded. This ion makes up the conjugate base of nitric acid.
Explanation:
Answer:
The Order is as follow,
C-H < S-H < H-Br < H-Cl
Explanation:
Polarity depends on the electronegativity difference between two atoms, greater the electronegativity difference, greater will be the polarity of bond and vice versa.
Electronegativity Difference between Hydrogen and other given elements are as follow,
1) C-H;
E.N of Carbon = 2.55
E.N of Hydrogen = 2.20
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Difference 0.35
2) S-H;
E.N of Sulfur = 2.58
E.N of Hydrogen = 2.20
------------
Difference 0.38
3) H-Br;
E.N of Bromine = 2.96
E.N of Hydrogen = 2.20
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Difference 0.76
4) H-Cl;
E.N of Chlorine = 3.16
E.N of Hydrogen = 2.20
-----------
Difference 0.96
Hence it is proved that the greatest electronegativity difference is found between H and Chlorine in H-Cl, therefore it is highly polar bond and vice versa.
Answer:
THE EMPIRICAL FORMULA FOR THE UNKNOWN COMPOUND IS C7H9O
Explanation:
The empirical formula for the unknown compound can be obtained by following the processes below:
1 . Write out the percentage composition of the individual elements in the compound
C = 75.68 %
H = 8.80 %
O = 15.52 %
2. Divide the percentage composition by the atomic masses of the elements
C = 75 .68 / 12 = 6.3066
H = 8.80 / 1 = 8.8000
O = 15.52 / 16 = 0.9700
3. Divide the individual results by the lowest values
C = 6.3066 / 0.9700 = 6.5016
H = 8.8000 / 0.9700 = 9.0722
O = 0.9700 / 0.9700 = 1
4. Round up the values to the whole number
C = 7
H = 9
O = 1
5 Write out the empirical formula for the compound
C7H90
In conclusion, the empirical formula for the unknown compound is therefore C7H9O
Answer:
1.98x10⁻¹² kg
Explanation:
The <em>energy of a photon</em> is given by:
h is Planck's constant, 6.626x10⁻³⁴ J·s
c is the speed of light, 3x10⁸ m/s
and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)
- E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J
Now we multiply that value by <em>Avogadro's number</em>, to <u>calculate the energy of 1 mol of such protons</u>:
- 1 mol = 6.023x10²³ photons
- 2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J
Finally we <u>calculate the mass equivalence</u> using the equation:
- m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg
