Answer: Atomic mass of gallium = 69.72304093 amu
Explanation:
We calculate the atomic mass of Gallium by using the Relative abundance of the isotopes.
Atomic mass =mass x Natural Abundance of Isotope 1 +mass x Natural Abundance of Isotope 2
(60.11/100 x 68.9256) + (39.89/100 x 70.9247)
= 41.43117816 + 28.29186283 = 69.72304093 amu
Answer: % Errror = 11.05%
Explanation:
Percentage error is the difference between the measured value and the theoretical value as expressed in percent.
That is;
% Error =
( Theoretical value -measured value) /(Theoretical value) × 100
Theoretical value =55.2kJ
Measured value = 49.1kJ
% Error = (55.2-49.1)/(55.2) ×100
%Error = 6.1/55.2 ×100
% Error = 0.1105 ×100
% Error = 11.05
Therefore the percentage error is 11.05%
The answer is B. Lithosphere
Explanation: The upper layer of mantle is a zone of rigid brittle rock. Below the lithosphere is asphalt-like and is called the Asthenosphere.
Answer:
Explanation:
In a chemical formula, the oxidation state of transition metals can be determined by establishing the relationships between the electrons gained and that which is lost by an atom.
We know that for compounds to be formed, atoms would either lose, gain or share electrons between one another.
The oxidation state is usually expressed using the oxidation number and it is a formal charge assigned to an atom which is present in a molecule or ion.
To ascertain the oxidation state, we have to comply with some rules:
- The algebraic sum of all oxidation numbers of an atom in a neutral compound is zero.
- The algebraic sum of all the oxidation numbers of all atoms in an ion containing more than one kind of atom is equal to the charge on the ion.
For example, let us find the oxidation state of Cr in Cr₂O₇²⁻
This would be: 2x + 7(-2) = -2
x = +6
We see that the oxidation number of Cr, a transition metal in the given ion is +6.
Here we have to get the number of strontium atom needed to be laid side by side to span a distance of 2.10 mm.
9.03×10⁷ number of strontium atoms are needed to cover the distance of 2.10mm.
The radius of the strontium atom is 215 pm, thus the diameter of strontium atom is (2×215) = 430 pm.
Now we know 1 pm is equivalent to 10⁻⁹ mm. Thus 430 pm is equivalent to (430×10⁻⁹) = 4.3×10⁻⁷ mm.
Thus 1 strontium atom can occupy 4.3×10⁻⁷mm of space. Now the distance to be occupied by the lay down of the strontium atom side by side of the span is 2.10 mm.
Hence, 
×10⁷ = 9.03×10⁷ number of atoms are needed.
