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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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What was Thomson's model of the atom called?
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3 0
3 years ago
I need help what is s-1/3=7/9???? please help guys and thank you
nalin [4]

Answer:

10/9

Explanation:

First, let's convert 1/3 and 7/9 so that the have the same denominator. To do this let's find the least common multiple of 3 and 9.

List the multiples of 3 and 9:

3: 3, 9

9: 9

They have a least common multiple of 9

We need to convert 1/3 so it has a denominator of 9:

1/3*3/3 (we can multiply it by 3/3 because any number over itself is 1) = 3/9

s-3/9=7/9

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6 0
2 years ago
A coffee-cup calorimeter initially contains 125 g water at 24.28C. Potassium bromide (10.5 g), also at 24.28C, is added to the w
iren2701 [21]

Answer:

The solution is given below

Explanation:

Heat, q= mc∆T

q= 125g x 4.18 J/g∙°C x (21.18x- 24.28) °C

q=  -1619.75J

NEGATIVE SIGN INDICATES THAT HEAT IS ABSORBED.

Enthalpy Change, ∆H = 1619.75 7/ 10.5 g

                                     = 154.26 J/g

No. of moles of KBr = Mass of KBr/ Molecular Weight of KBr

                                =10.5g/119gmol-1

                                =0.088 mol

∆H= 1619.75 J/ 0.088 mol

      = 18.41 kJ/mol  

6 0
3 years ago
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