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sweet [91]
3 years ago
6

Classify each substance as a strong acid, strong base, weak acid, or weak base. drag each item into the appropriate bin.

Chemistry
1 answer:
Zepler [3.9K]3 years ago
6 0
Since you forgot to include the choices for classification, I would just define each of these and tell you the hints that would help you classify them.

Among these acids and bases, its is the strong acids and strong bases that are easily classified. You should note that there are only 7 strong acids existing. All the rest are weak acids. These 7 acids are: HCl, HBr, HI, HClO₃, HClO₄, HNO₃ and H₂SO₄. On the other hand, there are only 8 strong bases; the rest are weak bases. These are the hydroxides of the Group ! and !! metals: LiOH, NaOH, KOH, RbOH, CsOH, Ca(OH)₂, Sr(OH)₂, and Br(OH)₂.

For the weak acids and weak bases, just remember the definitions of Arrhenius, Lewis and Bronsted-Lowry. A weak base are those compounds that accept H⁺ protons, produce OH⁻ ions when solvated and an electron donor. A weak acid are those compounds that donate H⁺ protons, produce H⁺ ions when solvated and an electron acceptor.
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In chemistry, attention is the abundance of a constituent divided by way of the total quantity of an aggregate. several types of mathematical descriptions may be prominent: mass concentration, molar concentration, range concentration, and quantity concentration.

it's miles the amount of solute dissolves in one hundred g solvent. If the attention of the answer is 20 %, we understand that there are 20 g solutes in one hundred g solution. instance: 10 g salt and 70 g water are mixed and the solution is ready. find awareness of the answer by means of percentage mass.

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1 year ago
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Anna71 [15]

Based on the dilution formula, 0.1 mL of the stock solution of the enzyme is required to prepare a 50-fold diluted enzyme in 0.01 M HCl.

<h3>How can 50-fold dilution of the enzyme be done?</h3>

The 50-fold dilution of the stock enzyme solution can be done by using the dilution formula to determine the given volume of the stock solution required.

The dilution formula is given below:

  • C1V1 = C2V2

where:

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  • C2 = Final concentration of enzyme
  • V1 = Initial volume
  • V2 = Final volume

From the given data for the enzyme dilution;

C1 = 1

C2 = 1/50 = 0.02

V1 = x

V2 = 5 ml

Making V1 subject of formula in the dilution formula:

V1 = C2V2/C1

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Therefore, 0.1 mL of the stock solution of the enzyme is required to prepare a 50-fold diluted enzyme in 0.01 M HCl.

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7 0
2 years ago
If 42.8 mL of 0.204 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solut
Aneli [31]

Hey There!

At neutralisation moles of H⁺ from HCl  = moles of OH⁻ from Ca(OH)2  so :

0.204 * 42.8 / 1000  => 0.0087312 moles

Moles of Ca(OH)2 :

2 HCl + Ca(OH)2 = CaCl2 + 2 H2O

0.0087312 / 2 => 0.0043656 moles (  since each Ca(OH)2 ives 2 OH⁻ ions )

Therefore:

Molar mass Ca(OH)2 = 74.1 g/mol

mass = moles of Ca(OH)2 * molar mass

mass =  0.0043656 * 74.1

mass = 0.32 g of Ca(OH)2


Hope that helps!

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