The image above shows what happens when hydrochloric acid (HCl) and sodium hydroxide (NaOH) are mixed with water. Which statemen
t below is true about both solutions?
A. Both solutions contain solutes that do not dissolve completely in water.
B. Both solutions are saturated because they contain the maximum concentration of a solute dissolved in the solvent.
C. Both solutions will conduct an electric current because they both contain ions that can carry a charge.
D. Both solutions are acidic because they both contain hydrogen.
A. Both solutions contain solutes that do not dissolve completely in water.
B. Both solutions are saturated because they contain the maximum concentration of a solute dissolved in the solvent.
C. Both solutions will conduct an electric current because they both contain ions that can carry a charge.
D. Both solutions are acidic because they both contain hydrogen.
1 answer:
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540g
Explanation:
Given parameters:
Mass of the aluminium oxide = 1020g
Unknown:
Mass of aluminium = ?
Solution:
To find the mass of the aluminium formed from this reaction, we work from the known to the unknown. The known here is the mass of the aluminium oxide.
Using this mass, find the number of moles in the aluminium and relate it using the balanced equation to that of the unknown aluminium.
From the number of moles, we can easily find the mass of the aluminium.
Solving:
Balanced equation:
2Al₂O₃ → 4Al + 3O₂
Number of moles of Al₂O₃ =
Molar mass of Al₂O₃ = 2(27) + 3(16) = 102g/mol
Number of moles = = 10mol
From the balanced equation:
2 moles of Al₂O₃ produced 4 moles of Al
10 moles of Al will produce = 20moles of Al
Mass of Al = number of moles of Al x molar mass of Al = 20 x 27 = 540g
Learn more:
Number of moles brainly.com/question/13064292
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Answer:
0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.
Explanation:
Volume of the radiator = 3.6 L
Percentage of antifreeze = 60%
Total volume of anti freeze in radiator = 60% of 3.6 L :
Percentage of water= 40%
Given,optimal cooling of the engine is obtained with only 50% antifreeze.
So, now we want to reduce the percentage of antifreeze from 60% to 50 %
Volume of coolant removed = x
Volume of water added = x
Volume of anti freeze removed = 60% of(x) = 0.6x
Volume of antifreeze left in radiator =50% of 3.6 L =
1.8 Liter is the desired volume of the antifreeze.
Total volume - Removed volume = desired volume
2.16 L - 60% of( x) = 1.8 L
0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.