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katrin2010 [14]
1 year ago
14

What are the relative numbers of h3o+ and oh- ions in an acidic and alkaline and a neutral solution?

Chemistry
2 answers:
Kryger [21]1 year ago
7 0

Explanation:

  • When a solution is acidic in nature then it means its pH range is from 1 to 6.9. And, in an acidic solution there are more number of hydrogen (H^{+}) or hydronium (H_{3}O^{+}) ions as compared to hydroxide ions (OH^{-}).

    For acidic solutions,    [H^{+}] or [H_{3}O^{+}] > [OH^{-}]

  • Whereas if a solution is neutral in nature then its pH will be equal to 7. This means that concentration of hydrogen ions will be equal to concentration of hydroxide ions.

    For neutral solutions,    [H^{+}] or [H_{3}O^{+}] = [OH^{-}]

  • If a solution is basic in nature then its pH will be greater than 7. This means that in a basic solution concentration of hydroxide ions is more than the concentration of hydrogen ions.

    For basic solutions,   [H^{+}] or [H_{3}O^{+}]  < [OH^{-}]

lana [24]1 year ago
3 0
Acid more H3O+ than OH-
Base less H3O+ than OH-
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Answer:

The essence including its particular subject is outlined in the following portion mostly on clarification.

Explanation:

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1 year ago
What type of reaction is the digestion of solid copper wire by nitric acid?
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Copper nitrate and nitric oxide are produced in this reaction.

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The data in the table below were obtained for the reaction: 2clo2 (aq) + 2 oh- (aq) --&gt; clo3- (aq) + clo2- (aq) + h2o (l) exp
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Lets organise the data given in the question
                [ClO₂] (m)       [OH⁻] (m)        initial rate (m/s)
                  <span>0.060              0.030               0.0248
</span><span>                  0.020              0.030               0.00276
</span><span>                  0.020              0.090                0.00828
rate equation as follows 
rate = k [</span>ClO₂]ᵃ [OH⁻]ᵇ
where k - rate constant 
we need to find order with respect to ClO₂ therefore lets take the 2 equations where OH⁻ is constant.
1) 0.00276 = k [0.020]ᵃ[0.030]ᵇ
2) 0.0248 = k [0.060]ᵃ[0.030]ᵇ
divide first equation from the second
0.0248/0.00276 = [0.060/0.020]ᵇ
8.99 = 3ᵇ
8.99 rounded off to 9
9 = 3ᵇ
b = 2
order with respect to ClO₂ is 2
3 0
1 year ago
Write a balanced equation for the transmutation that occurs when a scandium-48 nucleus undergoes beta decay.
Tju [1.3M]

Answer:

A scandium-48 nucleus undergoes beta-minus decay to produce a titanium-48 nucleus.

\rm ^{48}_{21}Sc \to ^{48}_{22}Ti + ^{\phantom{1}\,0}_{-1}e^{-} + \bar{\mathnormal{v}}_e.

Explanation:

There are two types of beta decay modes: beta-minus and beta-plus.

In both decay modes, the mass number of the nucleus stays the same.

However, in a beta-minus decay, the atomic number of the nucleus increases by one. In a beta-plus decay, the atomic number decreases by one.

Each beta-minus decay releases one electron and one electron antineutrino. Each beta-plus decay releases one positron and one electron neutrino.

Look up the atomic number and relative atomic mass for the element scandium.

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This question did not specify whether the decay here is beta-plus or a beta-minus. However, the relative atomic mass of this element can give a rough estimate of the mode of decay.

Each element (e.g, \rm Sc) can have multiple isotopes. These isotopes differ in mass. The relative atomic mass of an element is an average  across all isotopes of this element. This mass is weighted based on the relative abundance of the isotopes. Its value should be closest to the most stable (and hence the most abundant) isotope.

The mass number of scandium-48 is significantly larger than the relative atomic mass of this element. In other words, this isotope contains more neutrons than isotopes that are more stable. There's a tendency for that neutron to convert to a proton- by beta-minus decay, for example.

The atomic number of the nucleus will increase by 1. 21 + 1 = 22. That corresponds to titanium. The mass number stays the same at 48. Hence the daughter nucleus would be titanium-48. Note that two other particles: one electron and one electron \rm e^{-} and one antineutrino \bar{v}_{\text{e}} (note the bar.) The neutrino helps balance the lepton number of this reaction.

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