I believe the correct answer from the choices listed above is option B. When equal masses of each are added to water, ethylene glycol would be more effective. Ethylene glycol is the most widely used automotive cooling-systemantifreeze, although methanol, ethanol, isopropyl alcohol, and
propylene glycol are also used.
Hope this answers the question. Have a nice day.
The amount, in mL, of the concentrated acid required, would be 1.1875 mL
<h3>Dilution</h3>
From the dilution equation:
m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.
m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL
v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL
Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.
More on dilution can be found here: brainly.com/question/13949222
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Answer:
yes 4K + O2 ------>2K20 is true.
