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guapka [62]
2 years ago
6

What happens to the conductivity of water when an acid is dissolved in it for bases?

Chemistry
1 answer:
Degger [83]2 years ago
8 0
An Arrhenius acid is therefore any substance that ionizes when it dissolves<span> in </span>water to give the H+<span>, or hydrogen, ion. An Arrhenius </span>base<span> is any substance that gives the OH</span>-<span>, or hydroxide, ion when it </span>dissolves<span> in </span>water<span>.</span>
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Referring to an activity series, which of the following combinations of reactants would not produce a successful single-replacem
loris [4]
 I believe the answer is Ca + MgSO4
8 0
3 years ago
Read 2 more answers
Can someone explain this to me, please?
zimovet [89]

Answer:

Isotopes have same atomic numbers, no. of protons and no. of electrons. Only their no. of neutrons and atomic mass are changed.

<u>Na - 24:</u>

Atomic Mass = 24

Atomic No. = 11

Hence,

No. of protons in Na-24 = 11

No. of neutrons = Atomic Mass - Atomic Number

No. of neutrons = 24 - 11

No. of neutrons = 13

Atomic Number = 11

\rule[225]{225}{2}

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>
5 0
2 years ago
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ikadub [295]

Answer:

A.

Explanation:

hope it helps...

4 0
2 years ago
Calcium forms ions with a charge of +2. Iodine forms ions with a charge of -1. Which of the following would represent an ionic c
astra-53 [7]

 The   formula that would represent an ionic  compound  that is composed of calcium and iodide ions  is   CaI2

 Explanation

Ionic compound  CaI₂ is formed  when Calcium form cation ( <em>a positively charged ion</em>) by losing 2 electrons while two iodine atoms form  anion ( <em>a negatively charged ion</em>)  by  gaining one electron each.

When writing down   formula of ionic compound, the formula  of cation  is written  first followed by anion  formula. therefore  Ca is written  first  followed by I.

The numeric subscript 2 after I(iodine) indicate that 2 atoms of iodine are involved  in bonding.

5 0
3 years ago
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
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