Name each of the following species for the following acid-base reactions. (The equilibrium lies to the right in each case, i.e.,
1 answer:
Answer: a)
acid : hydronium ion
base : methoxide ion
conjugate acid : methanol
conjugate base: water
b)
acid : hydrogen chloride
base : ethoxide ion
conjugate acid : ethanol
conjugate base: chloride ion
c)
acid : methanol
base : amide ion
conjugate acid : ammonia
conjugate base: methoxide ion
Explanation:
According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.
The species accepting a proton is considered as a base and after accepting a proton, it forms a conjugate acid.
The species losing a proton is considered as an acid and after loosing a proton, it forms a conjugate base
For the given chemical equation:
a)
acid : hydronium ion
base : methoxide ion
conjugate acid : methanol
conjugate base: water
b)
acid : hydrogen chloride
base : ethoxide ion
conjugate acid : ethanol
conjugate base: chloride ion
c)
acid : methanol
base : amide ion
conjugate acid : ammonia
conjugate base: methoxide ion
.
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Answer:
13.85 kJ/°C
-14.89 kJ/g
Explanation:
<em>At constant volume, the heat of combustion of a particular compound, compound A, is − 3039.0 kJ/mol. When 1.697 g of compound A (molar mass = 101.67 g/mol) is burned in a bomb calorimeter, the temperature of the calorimeter (including its contents) rose by 3.661 °C. What is the heat capacity (calorimeter constant) of the calorimeter? </em>
<em />
The heat of combustion of A is − 3039.0 kJ/mol and its molar mass is 101.67 g/mol. The heat released by the combustion of 1.697g of A is:
According to the law of conservation of energy, the sum of the heat released by the combustion and the heat absorbed by the bomb calorimeter is zero.
Qcomb + Qcal = 0
Qcal = -Qcomb = -(-50.72 kJ) = 50.72 kJ
The heat capacity (C) of the calorimeter can be calculated using the following expression.
Qcal = C . ΔT
where,
ΔT is the change in the temperature
Qcal = C . ΔT
50.72 kJ = C . 3.661 °C
C = 13.85 kJ/°C
<em>Suppose a 3.767 g sample of a second compound, compound B, is combusted in the same calorimeter, and the temperature rises from 23.23°C to 27.28 ∘ C. What is the heat of combustion per gram of compound B?</em>
Qcomb = -Qcal = -C . ΔT = - (13.85 kJ/°C) . (27.28°C - 23.23°C) = -56.09 kJ
The heat of combustion per gram of B is:
Answer:
1s22s22p6: Neon (Ne)
1s22s22p63s23p3: Phosphorous (P)
1s22s22p63s23p64s1: Potassium (K)
1s22s22p63s23p64s2(im not sure what 308 is supposed to be): Calcium (Ca)
1s22s22p63s23p64s23d104p65s24d3: there is no pure element that ends 4d3 that I know of so this can either be Zirconium(Zr) if it ends in 4d2 or Niobium (Nb) if it ends in 4d4
Explanation:
you can look at the periodic table and the trends to find the rough idea of where the electron configuration ends, there are helpful articles and images on these, i attached an image that may help. After that you can look at the atomic number to find the number of electrons for a pure element and use the electron subshell pattern thing to find the exact number
