Cross-section of a metal wire wrapped in plastic insulation the metal wire is an element.
<h3>What is an element?</h3>
An element is a substance or material that cannot be affected by an external substance or cannot be broken down by any reaction.
Metal wire is serving as an element, it should not be easily broken or penetrated as an insulator.
Therefore, Cross-section of a metal wire wrapped in plastic insulation the metal wire is an element.
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Particle mass charge location
protons 1.673*10 ^ - 27 kg 1.6*10 ^ -19 C in the nucleus
neutrons 1.675*10 ^ - 27 kg 0 in the nucleus
electron 9.11 * 10 ^ - 31 kg -1.6 * 10 ^ - 19 C around the nucleus (orbitals)
Protons and neutrons have almost same masses. Mass of electrons is 1/1840 the mass of the protons.
Protons and electrons have the same magnitud of charge with different sign. Protons are positive and electrons are negative. Neutrons do not have charge.
Answer:
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Explanation:
<u>Step 1: </u>Data given
The solution contains 0.036 M Cu2+ and 0.044 M Fe2+
Ksp (CuS) = 1.3 × 10-36
Ksp (FeS) = 6.3 × 10-18
Step 2: Calculate precipitate
CuS → Cu^2+ + S^2- Ksp= 1.3*10^-36
FeS → Fe^2+ + S^2- Ksp= 6.3*10^-18
Calculate the minimum of amount needed to form precipitates:
Q=Ksp
<u>For copper</u> we have: Ksp=[Cu2+]*[S2-]
Ksp (CuS) = 1.3*10^-36 = 0.036M *[S2-]
[S2-]= 3.61*10^-35 M
<u>For Iron</u> we have: Ksp=[Fe2+]*[S2-]
Ksp(FeS) = 6.3*10^-18 = 0.044M*[S2-]
[S2-]= 1.43*10^-16 M
CuS will form precipitates before FeS., because only 3.61*10^-35 M Sulfur Ions are needed for CuS. For FeS we need 1.43*10^-16 M Sulfur Ions which is much larger.
The precipitate is CuS.
Sulfide will precipitate at [S2-]= 3.61*10^-35 M
Answer: -
71
Explanation: -
From the diagram, we see that volume goes from 70 to 75 in 5 markings.
Each marking is for 1.
It is the most accuracy possible as there is no smaller marking.
Significant figures expresses the required amount of accuracy.
Thus the volume indicated from the diagram is 71 .
