The ore contains 55.4% calcium phosphate (related to the mineral apatite) so the amount of Ca3(PO4)2 is 55.4%x=1000g so x=1000/0.554= 1.805kg. Now for the % of P in this amount of calcium phosphate, use all the masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (NB oxygen is 16 mass x 4 =64) so the total mass is 310.2 and we have 61.95 of P (Pmass x 2) so 61.95/3102.= 0.19 or 19% P. So of the 1.805 x 0.19= 0.34kg of phosphorus.
the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment
theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations
we need to first find the theoretical yield
2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂
molar ratio of Na₂O₂ to Na₂CO₃ is 2:2
number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed
number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol
therefore number of Na₂CO₃ moles formed is - 0.10 mol
mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g
therefore theoretical yield is 10.6 g
percent yield = actual yield / theoretical yield x 100%
81.0 % = actual yield / 10.6 g x 100 %
actual yield = 10.6 x 0.81
actual yield = 8.59 g
therefore actual yield is 8.59 g
<h3>The enthalpy of combustion per mole of anthracene : 7064 kj/mol(- sign=exothermic)</h3><h3>Further explanation </h3>
The law of conservation of energy can be applied to heat changes, i.e. the heat received/absorbed is the same as the heat released
Q in = Q out
Heat can be calculated using the formula:
Q = mc∆T
Heat released by anthracene= Heat absorbed by water
Heat absorbed by water =
mol of anthracene (MW=178,23 g/mol)
The enthalpy of combustion per mole of anthracene :
Answer:
QED, EDQ, DQE, EQD, DEQ, QDE
Explanation:
The structure of all possible peptides that contain the given amino acids are :
QED, EDQ, DQE, EQD, DEQ, QDE
where : Asp is represented by the letter code D
Glu is represented by the letter code E
Gln is represented by the letter code Q
Note : when three amino acids combine they form what is known as tripeptide ( i.e. contains two peptide linkages ) while a peptide linkage is been formed by the combination of a carboxyl group of an amino acid and the amino group of different amino acid
[OH⁻] = 1.6 × 10⁻⁸ mol / dm³
<h3>Explanation</h3>
By definition, ![[\text{H}^{+}] = 10^{-\text{pH}}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D)
, where ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is the concentration of proton in the solution.
pH = 6.2 for this solution. As a result, 
.
![[\text{H}^{+}] \cdot [\text{OH}^{-}] = \text{K}_w](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Ctext%7BK%7D_w)
, where ![[\text{OH}^{-}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
the concentration of hydroxide ions and 
is the dissociation constant of water.
at 0.10 MPa and 25 °C. As a result, ![[\text{OH}^{-}] = \text{K}_w / [\text{H}^{+}] = 10^{14} / (6.31 \times 10^{-7}) = 1.6 \times 10^{-8} \; \text{mol}\cdot \text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Ctext%7BK%7D_w%20%2F%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B14%7D%20%2F%20%286.31%20%5Ctimes%2010%5E%7B-7%7D%29%20%3D%201.6%20%5Ctimes%2010%5E%7B-8%7D%20%5C%3B%20%5Ctext%7Bmol%7D%5Ccdot%20%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
