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sashaice [31]
3 years ago
8

Despite the superior intellect of our lab workers, none of the problems above actually caused the discrepancies. This occurs som

etimes with unexpected or unexplained results, where multiple hypotheses can explain the cause of the discrepancy occur. Here is a challenge: closely examine each video, and see if you are more observant than our lab workers by determining the actual source of the error.
Chemistry
1 answer:
Soloha48 [4]3 years ago
6 0

Answer:

Haven't evaporated all of the water

Explanation:

One of the main sources of error that occur in a formula of a hydrate lab is that all of the water is not evaporated. We can see at the end of the video that half of the CoCl2 is a light blue colors and the other half is a dark blue color. This indicates that all of the water still has not been evaporated off, resulting in the actual mass of the salt to be greater than the predicted value.

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for what

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What is the [H+] and [OH-] of a 3.5 M HCIO3 solution?
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2 years ago
if 3.26 g of FeNO33 is dissolved in enough water to make exactly what is the molar concentration of nitrate ion g
Tanzania [10]

Answer:

0.404M

Explanation:

...<em>To make exactly 100.0mL of solution...</em>

Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.

To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:

<em>Moles Fe(NO₃)₃ -Molar mass: 241.86g/mol-:</em>

3.26g * (1mol / 241.86g) =

0.01348 moles Fe(NO₃)₃ * (3 moles of NO₃⁻ / 1mole Fe(NO₃)₃) =

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5 0
2 years ago
Write a balanced half-reaction describing the reduction of aqueous vanadium(V) cations to aqueous vanadium(I) cations.
11Alexandr11 [23.1K]

Explanation:

Reduction is a chemical reaction in which electrons are gained by one of the atoms taking part in the reaction and lowering of an oxidation state of that atom.

Reduction takes place at the cathode.

In aqueous, vanadium(V) is present in +5 oxidation state which on reduction changes to vanadium(I) with +1 oxidation state.

The half reaction is :

V^{5+}(aq)+4e^- \rightarrow V^{1+}(aq)


5 0
3 years ago
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