Answer: All of the statements are true.
Explanation:
(a) Considering the system mentioned in the equation:-
The sum of total moles in the flask will always be equal to 1 which leads to confirmation of this statement as for 60 secs= 0.16 mol A and 0.84 mol B
(b) 0<t< 20s, mole A got reduced from 1 mole to 0.54 moles while at 40s to 60s A got decreased from 0.30 moles to 0.16 moles.
0 to 20s is 0.46 (1 - 0.54 = 0.46)mol whereas,
40 to 60s is 0.14 (0.30-.16 = 0.14) mol
(0.46 > 0.14) mol leading this statement to be true as well.
(c) Average rate from t1 = 40 to t2 = 60 s is given by:
which is true as well
Answer:
1. Galvanic oxidation. Example is the corrosion of aluminium wires when in contact with copper wires under wet conditions.
2. Rainwater or Damp/moist air
3. Chromium-plated steel screws or stainless steel screws or galvanized steel screws
Explanation:
1. Galvanic oxidation or corrosion occurs when two different metals with different electrode potentials are brought into contact with each other by means of an electrolyte (usually a aqueous solution), such that a redox reaction occurs leading to one metal with the more negative electrode potential (the anode) becoming oxidized, while the other less negative potential (the cathode) is reduced.
In order for galvanic corrosion to occur, three elements are required.
i. Two metals with different corrosion potentials (anode and cathode)
ii. Direct metal-to-metal electrical contact
iii. A conductive electrolyte solution (e.g. water) must connect the two metals on a regular basis.
For example oxidation (corrosion) of aluminium wires when in contact with copper wire under wet conditions.
2. The most likely electrolyte will be rainwater containing dissoved solutes (if the panel is in an exposed part of the house) or damp/moist air.
3. From the table, the most likely screw will be chromium-plated steel screws or stainless steel (made of iron and nickel) screws or galvanized steel (zinc-plated) screws.
All these possible screw components have a more negative electrode potential than copper. Thus they will serve as the anode in a galvanic oxidation with copper.
<span>The half-life of a first-order reaction is determined as follows:
</span>t½<span>=ln2/k
From the equation, we can calculate the </span><span>first-order rate constant:
</span>k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³
When we know the value of k we can then calculate concentration with the equation:
A₀ = 2 g/100 mL
t = 2.5 h = 150min
A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml
= 6.3 × 10⁻⁴ mg / 100ml
Answer: Option (e) is the correct answer.
Explanation:
A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.
For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.
Whereas a bond that is formed by sharing of electrons is known as a covalent bond.
For example, 
will be a covalent compound as there is sharing of electron between carbon and bromine atom.
Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.
For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.
Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.
Answer D. Follow Le Chatelier's principle.
