The element which would have the lowest electronegativity is: an element with a small number of valence electrons and a large atomic radius.
Atomic radii can be defined as a measure of the size (distance) of the atom of a chemical element such as hydrogen, oxygen, carbon, nitrogen etc, typically from the nucleus to the valence electrons. The atomic radius of a chemical element decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon). Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).
Generally, atoms with relatively large atomic radii tend to have a low electronegativity, ionization energy and a low electron affinity.
Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Thus, number of valence electrons is typically used to determine the chemical properties of elements such as electronegativity.
Electronegativity can be defined as the ability or tendency of the atom of an chemical element to attract any shared pair of electrons.
In conclusion, a chemical element that has small number of valence electrons and a large atomic radius would have the lowest electronegativity.
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That is true because the warm air goes up and the cold air comes down
Answer:
CH₃CH₂CH₂COOH.
Explanation:
To know which option is correct, let us hydrolysed the given ester. This is illustrated in the attached photo.
Hydrolysis of ester involves breaking the ester bond by a water molecule to produce the corresponding alcohol and carboxylic acid.
From the reaction given in the attached photo, we can see that the carboxylic acid needed to produce the desired ester is butanoic acid, CH₃CH₂CH₂COOH.
Answer:
while atoms form together, they percentage their outermost electrons to create more sustainable strength states. This sharing bonds the atoms into an ionic shape or a molecule
Explanation:
i hope this help a little
Answer: 0.0944 gram of H2
Explanation:
Raising the T from 25 C (298 K) to 700 C (973 K) increases the pressure of each gas by:
2.0 atm x (973 K / 298 K) = 6.53 atm
Where
Kc = Kp because the moles of product equals the moles of reactants.
At equilibriuim, the amounts are
P(H2) = 6.53 - x
P(CO2) = 6.53 - x
P(H2O) = x
P(CO) = x
Kc = Kp = .534 = (x)(x) / [(6.53 - x)(6.53 - x)]
Take the square root of each side
(.534)^0.5 = x / (6.53 - x)
x = 0.731 (6.53 - x)
x = 4.77 - 0.731x
1.731x = 4.77
x = 4.77 / 1.731 = 2.76 atm
P(H2) at equilibriuim = 6.53 - 2.76 = 3.77 atm
P(CO2) at equilibrium = 6.53 - 2.76 = 3.77 atm
PV = nRT
n = PV/RT = [(3.77 atm)(1.00 L)] / [(0.08206 L atm/K mol)(973 K)] = 0.0472 mol H2
0.0472 mol H2 x (2.00 g / 1.00 mol) = 0.0944 g
