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8090 [49]
3 years ago
8

What is HI (aq), an acid, base, or neutral?

Chemistry
1 answer:
Leni [432]3 years ago
8 0
Answer: acid
please mark me brainliest
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Describe use of H₂S an analytical reagent?<br><br>​
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Answer:

The main use for hydrogen sulfide is in the production of sulfuric acid and elemental sulfur. ... H2S is used to prepare the inorganic sulfides you need to make those products. As a reagent and intermediate, hydrogen sulfide is beneficial because it can prepare other types of reduced sulfur compounds.

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What is the required molarity of a ba(oh)₂ solution to prepare a 1.0 m oh⁻ solution?
timurjin [86]
  The   molarity   of  a  Ba(OH)2  solution  required to prepare   a1.0 OH-  solution  is  calculated  as  follows


write   the  equation  for  dissociation of  Ba(Oh)2


that  is,

Ba(Oh)2  -----> Ba^2+   +  2Oh-

by  use  of  reacting   ratio  between  Ba(Oh)2  to  Oh  which  is   1:2  the  molarity of  Ba(oh)2  =    1.0/2 =  0.5 M
8 0
4 years ago
Read 2 more answers
52. In the following reaction, which are the Spectator ions? *
maks197457 [2]

Answer:

nitrate and sodium are spectator ions .The reactions that form an aqueous solution are often a spectator ions.

5 0
2 years ago
Two liquids that are soluble in each other in any proportion are said to be what?
Romashka-Z-Leto [24]
When two liquids are completely soluble in each other in all proportions, they are said to be miscible<span>. For example, ethanol and water are </span>miscible<span>. If the liquids do not mix, they are said to be </span>immiscible<span>.</span>
8 0
4 years ago
Telluric acid (H2TeH4O6) is a diprotic acid with Ka1 = 2.0x10-8 and Ka2 = 1.0x10-11. A 0.25 M H2TeH4O6 contains enough HCl so th
tensa zangetsu [6.8K]

Answer:

5x10⁻⁶ = [HTeH₄O₆⁺]

Explanation:

The first dissociation equilibrium of the telluric acid in water is:

H₂TeH₄O₆ + H₂O ⇄ HTeH₄O₆⁺ + H₃O⁺

Using H-H equation for telluric acid:

<em>pH = pKa + log₁₀ [HTeH₄O₆⁺] / [H₂TeH₄O₆]</em>

pKa of telluric acid is -logKa1

pKa = -log 2.0x10⁻⁸

pKa = 7.699

As concentration of [H₂TeH₄O₆] is 0.25M, replacing in H-H equation:

3.00 = 7.699+ log₁₀ [HTeH₄O₆⁺] / [0.25M]

-4.699 = log₁₀ [HTeH₄O₆⁺] / [0.25M]

2x10⁻⁵ = [HTeH₄O₆⁺] / [0.25M]

<h3>5x10⁻⁶ = [HTeH₄O₆⁺]</h3>

7 0
3 years ago
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