Answer:
The compound you will use is the Dibasic phosphate
Explanation:
Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,
In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.
To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.
O1Fl2
1. Assume an 100g sample, so the percentage will stay the same
2. Covert each element into their molar mass
29.6/16.00=1.8 mols of O
70.4/19.00=3.7 mols of Fl
3. Divide both by the smallest value of mol
1.8/1.8=1 O
3.7/1.8=2 Fl
4. Write the empirical formula:
O1Fl2
Answer:
instantaneous rate would be the term.
PbCr04 + P4O10
Hope this helps!
Answer:
Mass = 2.77 g
Explanation:
Given data:
Mass of HCl = 2 g
Mass of CaCl₂ produced = ?
Solution:
Chemical equation:
2HCl + Ca → CaCl₂ + H₂
Number of moles of HCl:
Number of moles = mass / molar mass
Number of moles = 2 g/ 36.5 g/mol
Number of moles = 0.05 mol
now we will compare the moles of HCl with CaCl₂.
HCl : CaCl₂
2 : 1
0.05 : 1/2×0.05 = 0.025 mol
Mass of CaCl₂:
Mass = number of moles × molar mass
Mass = 0.025 mol × 110.98 g/mol
Mass = 2.77 g
