Assuming the balloon initially has volume of 0 when deflated, the total P(deltaV) = (1.00 atm)(550,000 ft^3) = 550,000 atm-ft^3. To convert into work units, we can first convert ft^3 to L:
(550,000 atm-ft^3)(1 m/3.28 ft)^3
= (15,586.2 atm-m^3)
Then we convert to L:
(15,586.2 atm-m^3)(1000 L/m^3)
= 15,586,200 atm-L
Then we convert to J:
(15,586,200 atm-L)(101.325 J / 1 atm-L)
= 1.579 x 10^9 J
Answer:
4.8 %
Explanation:
We are asked the concentration in % by mass, given the molarity of the solution and its density.
0.8 molar solution means that we have 0.80 moles of acetic acid in 1 liter of solution. If we convert the moles of acetic acid to grams, and the 1 liter solution to grams, since we are given the density of solution, we will have the values necessary to calculate the % by mass:
MW acetic acid = 60.0 g/mol
mass acetic acid (the solute) = 0.80 mol x 60 g / mol = 48.00 g
mass of solution = 1000 cm³ x 1.010 g/ cm³ (1l= 1000 cm³)
= 1010 g
% (by mass) = 48.00 g/ 1010 g x 100 = 4.8 %
The average mass of an atom is calculated with the formula:
average mass = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2) + ... an so on
For the boron we have two isotopes, so the formula will become:
average mass of boron = abundance of isotope (1) × mass of isotope (1) + abundance of isotope (2) × mass of isotope (2)
We plug in the values:
10.81 = 0.1980 × 10.012938 + 0.8020 × mass of isotope (2)
10.81 = 1.98 + 0.8020 × mass of isotope (2)
10.81 - 1.98 = 0.8020 × mass of isotope (2)
8.83 = 0.8020 × mass of isotope (2)
mass of isotope (2) = 8.83 / 0.8020
mass of isotope (2) = 11.009975
mass of isotope (1) = 10.012938 (given by the question)
Explanation:
The - 3 degree C( carbon atom) 2p atomic orbital + methyl C-H sigma molecular orbital because one C-H bond has to dissolve its bond and provide the H that is sigma molecular orbital and the carbonation is type 3 degree sp2 carbon.
Hyperconjugation is the stabilizing effect arising from the electrons ' engagement in a π-bond (usually C-H or C-C) with a neighboring empty or partly filled p-orbital or π-orbital to provide an expanded molecular orbital that enhances system stability.
The ionization energy of an element is the amount of energy required to remove one mole of electrons from the element in its gaseous state. The equations for the first three are:
Fe(g) → Fe⁺(g) + e⁻
Fe⁺(g) → Fe⁺²(g) + e⁻
Fe⁺²(g) → Fe⁺³(g) + e
