What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC
l to reach the 2nd end point
1 answer:
Answer:
0.109 g.
Explanation:
Equation of the reaction:
Na3PO4 + 3HCl --> 3NaCl + H3PO4
Number of moles of HCl = molar concentration × volume
= 0.1 × 0.04
= 0.004 mol.
By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3
= 0.0013 mol
Mass of Na3PO4 = molar mass × number of moles
= 0.0013 × 164
= 0.219 g
Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance
= 0.219 × 50 g/100 g
= 0.109 g.
You might be interested in
Answer:
density = 8.9 g/ml
m = 20.5 g
V = ... ?
density = m/V
V = m / density
= 20.5 / 8.9
= 2.30337 ml
<span>https://www.onetonline.org/find/career?c=6</span>
Answer:
opinion b
Explanation:
the product of neutralization reaction between hcl and CA(oh)2 is option b.
Answer:
Hence the concentration of a MnO41- solution that has absorbance of 0.490 in the same cell at that wavelength is 0.3266.
Explanation:
Now A = el, el=const
Then,

Answer:
d = 0.93 g/cm³
Explanation:
Given data:
Mass of object = 28 g
Volume of object = 3cm×2cm×5cm
density of object = ?
Solution:
Volume of object = 3cm × 2cm ×5cm
Volume of object = 30 cm³
Density of object:
d = m/v
by putting values,
d = 28 g/ 30 cm³
d = 0.93 g/cm³