D is to stop the current and the force can be removed
C. Temperature should be the correct answer. Hope this helps, and Brainliest would be appreciated!
Answer is: dissolve 74,9 grams CuSO₄·5H₂O in one liter volumetric flask.
V(CuSO₄·5H₂O) = 1 L.
c(CuSO₄·5H₂O) = 0,30 mol/L.
n(CuSO₄·5H₂O) = V(CuSO₄·5H₂O) · c(CuSO₄·5H₂O) .
n(CuSO₄·5H₂O) = 1 L · 0,3 mol/L.
n(CuSO₄·5H₂O) = 0,3 mol.
m(CuSO₄·5H₂O) = n(CuSO₄·5H₂O) · M(CuSO₄·5H₂O).
m(CuSO₄·5H₂O) = 0,3 mol · 249,7 g/mol.
m(CuSO₄·5H₂O) = 74,9 g.
The answer is paper chromatography using different solvents with a range of polarities as the mobile phase.
Paper chromatography- Low-molecular-mass molecules can be separated using paper chromatography based on how evenly they are distributed in the stationary and mobile phases. Paper chromatography is regarded as a potent analytical technique because of its low cost and the availability of numerous procedures for the separation of chemicals.
A small amount of a sample solution is poured onto a strip of chromatography paper in a paper chromatography experiment. After that, a solvent is used to suspend the chromatography paper. The sample solution's constituent components split out into bands of distinct hue as the solvent goes up the paper.
The speed of the chromatography process is influenced by the solvent's polarity. Therefore, we may conclude that all of the other components in the mixture move more quickly during the chromatography experiment if the solvent's polarity is increased.
Thus, answer is paper chromatography using different solvents with a range of polarities as the mobile phase.
To learn more about paper chromatography refer- brainly.com/question/1394204
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Answer:
There is 1.6 L of NO produced.
Explanation:
I assume you have an excess of NH3 so that O2 is the limiting reagent.
<u>Step 1:</u> Data given
2.0 liters of oxygen reacts with ammonia
<u>Step 2:</u> The balanced equation
4NH3 + 5O2 → 4NO + 6H2O
For 4 moles of NH3, we need 5 moles of O2 to produce 4 moles of NO and 6 moles of H2O
Consider all gases are kept under the same conditions for pressure and temperature, we can express this mole ratio in terms of the volumes occupied by each gas.
This means: when the reaction consumes 4 liters of ammonia (and 5 liters of oxygen) it produces 4 liters of nitrogen monoxide
Now, when there is 2.0 liters of oxygen consumed, there is 4/2.5 = 1.6 L of nitrogen monoxide produced.
There is 1.6 L of NO produced.
