Answer:
Thus, 31 g potassium nitrate would be needed to produce a saturated solution of potassium nitrate in 50 g of water at 313 K. This is the maximum amount of solute that can be dissolved in water at 313K. The solution is saturated at that temperature. Mass KNO3 = 62/2 = 31g
Explanation:
Answer:
150 g/mol
Explanation:
Let's consider the complete neutralization of a diprotic acid H₂X with NaOH.
H₂X + 2 NaOH → Na₂X + 2 H₂O
40.0 mL of 0.200 M NaOH. were required to reach the endpoint. The reacting moles of NaOH are:
0.0400 L × 0.200 mol/L = 8.00 × 10⁻³ mol
The molar ratio of H₂X to NaOH is 1:2. The reacting moles of H₂X are 1/2 × 8.00 × 10⁻³ mol = 4.00 × 10⁻³ mol.
4.00 × 10⁻³ moles of H₂X have a mass of 0.600 g. The molar mass of H₂X is:
0.600 g/4.00 × 10⁻³ mol = 150 g/mol
Alkali Metals have 2 valence electrons
Answer:
21.1cm or 10.1cm
Explanation:
Since the percent error is 35.5% and the experimental measurement was 15.6cm, the two possible values for the actual measurement are 15.6 plus 35.5% of 15.6 or 15.6 minus 35.5% of 15.6.
This is this:
15.6 + (0.355 x 15.6)
= 15.6 + 5.54
= 21.1cm
Or this:
15.6 - (0.355 x 15.6)
= 15.6 - 5.54
= 10.1cm
