Answer is (4) - Se.
Among the given choices Se has the highest electronegativity value as 2.4 compared to others. Hence, Se shows <span>greatest attraction for electrons in a chemical bond.
</span>Electronegativity is
a value that tells us how an atom can attract electrons towards itself. <span>If
the electronegativity is high, then the attraction to the electrons is also high.
</span>
<span><span>Mn<span>O<span>2<span>(s)</span></span></span>+<span>H<span>2<span>(g)</span></span></span>→Mn<span>O<span>(s)</span></span>+<span>H2</span><span>O<span>(g)</span></span></span></span>
K + I - > KI
Potassium (needs to lose 1 electron) responds with Iodine (needs to pick up 1 electron) to fulfill both component's octet, shaping a salt, potassium iodide
This is a similar case for NaCl, simply unique components. Trust this made a difference.
Answer:
ΔH for formation of 197g Fe⁰ = 1.503 x 10³ Kj => Answer choice 'B'
Explanation:
Given Fe₂O₃(s) + 2Al⁰(s) => Al₂O₃(s) + 2Fe⁰(s) + 852Kj
197g Fe⁰ = (197g/55.85g/mol) = 3.527 mol Fe⁰(s)
From balanced standard equation 2 moles Fe⁰(s) => 852Kj, then ...
3.527 mole yield (a higher mole value) => (3.527/2) x 852Kj = 1,503Kj (a higher enthalpy value).
______
NOTE => If 2 moles Fe gives 852Kj (exo) as specified in equation, then a <u>higher energy value</u> would result if the moles of Fe⁰(s) is <u>higher than 2 moles</u>. The ratio of 3.638/2 will increase the listed equation heat value to a larger number because 197g Fe⁰(s) contains more than 2 moles of Fe⁰(s) => 3.527 mole Fe(s) in 197g. Had the problem asked for the heat loss from <u>less than two moles Fe⁰(s)</u> - say 100g Fe⁰(s) (=1.79mole Fe⁰(s)) - then one would use the fractional ratio (1.79/2) to reduce the enthalpy value less than 852Kj.
Answer:
The section of the bar is 2.92 inches.
Explanation:
Mass of the steel cut ,m = 1.00 kg = 1000 g
Volume of the steel bar = V = Area × height
Height of the of the section of bar = h
Area of Equilateral triangular = 
a = 2.50 inches
Cross sectional area of the steel mass = A


Density of the steel = d =




h = 2.92 inches
The section of the bar is 2.92 inches.