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1 year ago

Write an equation to show how HC2O4− can act as a base with HS− acting as an acid.

Chemistry

1 answer:

Artyom0805 [142]1 year ago

7 0

An acid donates $H^{+}$ ion in aqueous solution. A base accepts $H^{+}$ ion in aqueous solution.

The equation representing the acid base reaction of $HC_{2}O_{4}^{-}$ and $HS^{-}$ :

$HC_{2}O_{4}^{-}(aq) + HS^{-}(aq) ----> H_{2}C_{2}O_{4}(aq)+S^{2-}(aq)$

In the above reaction, as $HC_{2}O_{4}^{-}$ acts as a base it is accepting the hydrogen ion from $HS^{-}$ . Similarly, $HS^{-}$ donates its hydrogen ion to $HC_{2}O_{4}^{-}$ acting as an acid.

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All of the following reactions can be described as displacement reactions except:____________.

Lady_Fox [76]

Answer:

Explanation:

The reaction that is not a displacement reaction from all the options is $C_6H_6_{(l)} + Cl_{2(g)} --> C_6H_5Cl_{(l)} + HCl_{(g)}$

In a displacement reaction, a part of one of the reactants is replaced by another reactant. In single displacement reactions, one of the reactants completely displaces and replaces part of another reactant. In double displacement reaction, cations and anions in the reactants switch partners to form products.

Options a, c, d, and e involves the displacement of a part of one of the reactants by another reactant while option b does not.

Correct option = b.

8 0

1 year ago

A voltaic cell is constructed with two silver-silver chloride electrodes, where the half-reaction is AgCl (s) + e− → Ag (s) + Cl

ANTONII [103]

Answer : The cell emf for this cell is 0.118 V

Solution :

The half-cell reaction is:

$AgCl(s)+e^\rightarrow Ag(s)+Cl^-(aq)$

In this case, the cathode and anode both are same. So, $E^o_{cell}$ is equal to zero.

Now we have to calculate the cell emf.

Using Nernest equation :

$E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Cl^{-}{diluted}]}{[Cl^{-}{concentrated}]}$

where,

n = number of electrons in oxidation-reduction reaction = 1

$E_{cell}$ = ?

$[Cl^{-}{diluted}]$ = 0.0222 M

$[Cl^{-}{concentrated}]$ = 2.22 M

Now put all the given values in the above equation, we get:

$E_{cell}=0-\frac{0.0592}{1}\log \frac{0.0222M}{2.22M}$

$E_{cell}=0.118V$

Therefore, the cell emf for this cell is 0.118 V

4 0

1 year ago

Which of these did your answer include? High boiling and melting points: Hydrogen bonds increase the amount of energy required f

scZoUnD [109]

Here we have to get the right answers which include the given phrase.

The correct answers are as following:

High boiling and melting points: Hydrogen bond increase the amount of energy required for phase changes to occur, thereby raising the boiling and melting points.

High specific heat: Hydrogen bond increase the amount of energy required for molecules to increase the speed, thereby raising the specific heat.

High surface tension: Hydrogen bonds produce strong inter molecular attractions, which increase surface tension.

The incorrect answer:

Lower density as a solid than as a liquid: actually, density of solid is more than density of liquid as hydrogen bonds in solid produce strong inter molecular attractions among molecules, which aggregates molecules together, hence volume of associated molecules reduces. Therefore, density of solid is more than that of liquid.

7 0

1 year ago

Read 2 more answers

Draw all of the constitutional isomers of the molecule with formula C3H5Br. Ignore geometric and stereoisomers.

Kay [80]

Answer:

-) 3-bromoprop-1-ene

-) 2-bromoprop-1-ene

-) 1-bromoprop-1-ene

-) bromocyclopropane

Explanation:

In this question, we can start with the I.D.H (hydrogen deficiency index):

$I.D.H~=~\frac{(2C)+2+(N)-(H)-(X)}{2}$

In the formula we have 3 carbons, 5 hydrogens, and 1 Br, so:

$I.D.H~=~\frac{(2*3)+2+(0)-(5)-(1)}{2}~=~1$

We have an I.D.H value of one. This indicates that we can have a cyclic structure or a double bond.

We can start with a linear structure with 3 carbon with a double bond in the first carbon and the Br atom also in the first carbon (1-bromoprop-1-ene). In the second structure, we can move the Br atom to the second carbon (2-bromoprop-1-ene), in the third structure we can move the Br to carbon 3 (3-bromoprop-1-ene). Finally, we can have a cyclic structure with a Br atom (bromocyclopropane).

See figure 1

I hope it helps!