Is anyone good at chemistry if so can someone help me please ? (NO LINKS)

Chemistry

1 answer:

steposvetlana [31]3 years ago

4 0

In general, salts (formed during a neutralization reaction) are ionic compounds that are soluble in water and dissociate in solution into ions that conduct electricity. Out of the six statements given, there are three related statements that rehash the foregoing, and there are three related statements that are collectively incorrect.

Statements A, B, and D are (generally) true regarding salts formed during a neutralization reaction. When you consider that the net ionic equation of many acid-base neutralization reactions is H⁺(aq) + OH⁻(aq) → H₂O(l), the counterions of the H⁺(aq) and OH⁻(aq) are the aqueous spectator ions that comprise the salt. These ions are electrolytes, as they are charged species that can carry a current in solution; they are ionic compounds by definition since they're composed of cations and anions; and, as aqueous species, they're clearly dissolved in water.

Statements C, E, and F, as a whole, generally aren't true of such salts.

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pyrite is a mineral composed of 46.5 mass % iron and 53.5 mass % sulfur. determine the empirical formula for calcocite.

padilas [110]

The empirical formula for pyrite is FeS2.

HOW TO CALCULATE EMPIRICAL FORMULA:

The empirical formula represents the simplest whole number ratio of constituents element of a compound. The empirical formula of pyrite can be calculated as follows:

46.5 mass % Fe = 46.5g of Fe

53.5 mass % S = 53.5g of S

Next, we divide each element's mass value by its molar mass

Fe = 46.5g ÷ 56g/mol = 0.83mol

S = 53.5g ÷ 32g/mol = 1.67mol

Next, we divide each mole value by the smallest (0.83mol)

Fe = 0.83mol ÷ 0.83 = 1

S = 1.67mol ÷ 0.83 = 2.014

Approximately, the ratio of Fe to S is 1:2. Therefore, the empirical formula of pyrite is FeS2.

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3 0

2 years ago

For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia

Fittoniya [83]

Answer : The value of $E^o_{(Cu^{2+}/Cu)}$ is, 0.34 V

Explanation :

Here, copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

The oxidation-reduction half cell reaction will be,

Oxidation half reaction: $Zn\rightarrow Zn^{2+}+2e^-$

Reduction half reaction: $Cu^{2+}+2e^-\rightarrow Cu$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

The overall balanced equation of the cell is,

$Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu$