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Jlenok [28]
3 years ago
9

Which of the following best describes an alkali?

Chemistry
2 answers:
julia-pushkina [17]3 years ago
7 0

Answer:

A. Any substance with a pH above 7.

Explanation:

On the pH scale substances whose value is greater than 7 and less than 14 are termed alkali.

Acids have  a pH of less than 7 whereas neutral compounds such as water has pH of 7.

  • An alkali is any substance that produces excess hydroxyl ion in solutions.
  • Water soluble bases are known as alkali.
  • They are usually certain metallic oxides, metallic hydroxides and aqueous ammonia.
Vinil7 [7]3 years ago
6 0

Answer:

C   A soluble solid that forms H+ ions in a solution

Explanation:

founders ed

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Copper (II) nitrate, Cu(NO3)2, solution reacts with potassium hydroxide, KOH, to form a blue
Gekata [30.6K]

Answer:

Copper(II) nitrate and potassium hydroxide are soluble ionic compounds, which implies that they dissociate completely when dissolved in water to produce ions. ... You can thus say that the balanced chemical equation that describes this double ... Cu(NO3)2(aq)+2KOH(aq)→Cu(OH)2(s)⏐⏐↓+2KNO3(aq).

6 0
3 years ago
What is the energy of the photons of a laser with a frequency of 5.75 x 10^12 Hz? (Hint: Remember to round your answer to the lo
Simora [160]

Answer:

E = 3.81×10 ⁻²¹ J

Explanation:

Given data:

Frequency of photon = 5.75 ×10¹² Hz

Plancks constant = 6.626 ×10⁻³⁴ Js

Energy of photon = ?

Solution:

E = h×f

E = 6.626 ×10⁻³⁴ Js ×  5.75 ×10¹² s⁻¹

E = 38.1×10 ⁻²² J

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8 0
3 years ago
A 100.0 mL solution containing 0.923 gof maleic acid (MW=116.072 g/mol) is titrated with 0.265 M KOH. Calculate the pH of the so
Vlad1618 [11]

Answer:

pH = 9,57

[M²⁻] = 7,948x10⁻²M

[HM⁻] = 4x10⁻⁵M

[H₂M] = 0M

Explanation:

The moles of maleic acid presents in the solution are:

0,923g×\frac{1mol}{116,072g}=7,952x10⁻³moles of H₂M

60,0mL of 0,265M KOH are:

0,0600L×\frac{0,265mol}{1L}=0,0159 moles of KOH

The reactions of maleic acid (H₂M) and then with HM⁻ are:

H₂M + KOH → HM⁻ + H₂O + K⁺ (1)

HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)

For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:

0,0159 - 7,952x10⁻³ = <em>7,948x10⁻³ moles of KOH</em>

By (2), the moles produced of M²⁻ are the same as moles of KOH, <em>7,948x10⁻³  moles, </em>and moles of HM⁻ are:

7,952x10⁻³ -<em> </em>7,948x10⁻³ <em> = 4x10⁻⁶ moles of HM⁻</em>

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [M²⁻] /[HM⁻]

pH = 6,27 + log₁₀ <em>7,948x10⁻³ / 4x10⁻⁶</em>

pH = 9,57

The moles of M²⁻ are 7,948x10⁻³  and volume of the solution is 0,1000L,

[M²⁻] = 7,948x10⁻²M

Moles of HM⁻ are 4x10⁻⁶:

[HM⁻] = 4x10⁻⁵M

And there is not H₂M:

[H₂M] = 0M

I hope it helps!

8 0
3 years ago
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