Answer:
A feasible error could have been the removal of the sample before all water evaporated.
Explanation:
In order to determine the percentage of water in an hydrate, an experiment that could be performed is the heating of the sample until the mass does not change. If the student heated the sample an insufficient amount of time, water will be present in the sample, thus reducing the percentage reported.
Answer:
Empirical formula is CaSO₄.
Explanation:
Given data:
Percentage of calcium =29.44%
Percentage of sulfur = 23.55%
Percentage of oxygen = 47.01%
Empirical formula = ?
Solution:
Number of gram atoms of Ca = 29.44 / 40 = 0.74
Number of gram atoms of S = 23.55 / 32 = 0.74
Number of gram atoms of O = 47.01 / 16 = 3
Atomic ratio:
Ca : S : O
0.74/0.74 : 0.74/0.74 : 3/0.74
1 : 1 : 4
Ca : S : O = 1 : 1 : 4
Empirical formula is CaSO₄.
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Answer:
31.5 mL of a 2.50M NaOH solution
Explanation:
Molarity (M) is an unit of concentration defined as moles of solute (In this case, NaOH), per liter of solvent. That is:
Molarity = moles solute / Liter solvent
If you want to make 525mL (0.525L) of a 0.150M of NaOH, you need:
0.525L × (0.150mol / L) = <em>0.07875 moles of NaOH</em>
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If you want to obtain these moles from a 2.50M NaOH solution:
0.07875mol NaOH × (1L / 2.50M) = 0.0315L = <em>31.5 mL of a 2.50M NaOH solution</em>
Answer:
[OH-] = 6.17 *10^-10
Explanation:
Step 1: Data given
pOH = 9.21
Step 2: Calculate [OH-]
pOH = -log [OH-] = 9.21
[OH-] = 10^-9.21
[OH-] = 6.17 *10^-10
Step 3: Check if it's correct
pOH + pH = 14
[H+]*[OH-] = 10^-14
pH = 14 - 9.21 = 4.79
[H+] = 10^-4.79
[H+] = 1.62 *10^-5
6.17 * 10^-10 * 1.62 * 10^-5 = 1* 10^-14
