Answer:
Kindly check the attached picture for the diagram of the chemical compound.
Explanation:
So, the following parameters were given from the question above;
=> A triplet at 0.9 ppm and a quartet at 1.4 ppm, a singlet at 1.35 ppm. Now, the unknown compound has a molecular formula of C7H16O.
For a triplet at 0.9ppm, there are nine (9) atoms of hydrogen, for the quartet at 1.4ppm there are six(6) atoms of hydrogen and for the singlet at 1.35 ppm, the number of hydrogen atoms is one(1). Hence, the total number of hydrogen atoms = 16.
Therefore, number of bondings = [(2 × number of carbon atoms) + 2 - number of hydrogen atoms present on the compound)/2 .
Thus number of bonds =[( 2× 7) + 2 - 16] ÷ 2 = 0.
Hence, there is no double bond or ring in the compound.
Answer:
Phosphorous
Explanation:
Although elements such as gold, silver, tin, copper, lead and mercury have been known since antiquity, the first scientific discovery of an element occurred in 1649 when Hennig Brand discovered phosphorous.
Here is the correct question
You mix 125 mL of 0.170 M CsOH with 50.0 mL of 0.425 M HF in a coffee-cup calorimeter, and the temperature of both solutions rises from 20.20 °C before mixing to 22.17 °C after the reaction. What is the enthalpy of reaction per mole of ? Assume the densities of the solutions are all 1.00 g/mL, and the specific heat capacities of the solutions are 4.2 J/g · K. Enthalpy of reaction = kJ/mol
Answer:
75.059 kJ/mol
Explanation:
The formula for calculating density is:
Making mass the subject of the formula; we have :
mass = density × volume
which can be rewritten as:
mass of the solution = density × volume of the solution
= 1.00 g/mL × (125+ 50 ) mL
= 175 g
Specific heat capacity = 4.2 J/g.K
∴ the energy absorbed is = mcΔT
= 175 × 4.2 × (22.17 - 20.00) ° C
= 1594.95 J
= 1.595 J
number of moles of CsOH =
= 0.2125 mole
Therefore; the enthalpy of the reaction =
=
= 75.059 kJ/mol
Answer:
here is your answer...........
Answer:
Explanation:
HEPES is a zwitterion. That is, it has both the acid and base components in the same molecule. However, we can write its formula as HA. Then the equation for the equilibrium is
MM: 238.306
HA + H₂O ⇌ H₃O⁺ + A⁻; pKₐ = 7.56
m/g: 6.00
1. Calculate the moles of HEPES
2. Calculate the concentration ratio.
We can use the Henderson-Hasselbalch equation.
The acid and its conjugate base are in the same solution, so the concentration ratio is the same as the mole ratio.
3. Calculate the moles of HA and A⁻
4. Calculate the moles of KOH
We are preparing the buffer by adding KOH to convert HA to A⁻. The equation is
HA + OH⁻ ⟶ A⁻ + H₂O
The molar ratio is 1 mol A⁻:1 mol OH⁻, so we must use 0.009 742 mol of KOH.
5. Calculate the volume of KOH