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Licemer1 [7]
1 year ago
13

Why nitro group is called an ambident group?​

Chemistry
2 answers:
Zanzabum1 year ago
7 0

Answer:

The nitro group is an ambident group and is capable of getting attached to carbon chain through nitrogen. as well as through oxygen (-O - N = O) atom. The compound in which the -NO2 group is linked to the alkyl or aryl group through oxygen atom are called nitrites

podryga [215]1 year ago
4 0

Answer:

It can bond through either the N or O atoms  

Explanation:

An ambident ("both teeth") group is a group that can attach to another by either of two atoms.

A nitro group can bond through either N or O.

Organic chemistry

In organic chemistry, this property gives rise to functional group isomers.

For example, we can have either nitromethane (CH₃NO₂, Fig. 1) or methyl nitrite (CH₃ONO, Fig. 2)

Inorganic chemistry

The ambidentate property of the nitro group gives rise to linkage isomers in inorganic coordination complexes.

For example, we can have either a pentachloronitroiron(III) cation (Fig. 3) or a pentachloronitritoiron(III) cation (Fig. 4)

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A 0.500 g sample of C7H5N2O6 is burned in a calorimeter containing 600. g of water at 20.0∘C. If the heat capacity of the bomb c
Nata [24]

Answer:

22.7

Explanation:

First, find the energy released by the mass of the sample. The heat of combustion is the heat per mole of the fuel:

ΔHC=qrxnn

We can rearrange the equation to solve for qrxn, remembering to convert the mass of sample into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released by the reaction must be equal to the sum of the heat absorbed by the water and the calorimeter itself:

qrxn=−(qwater+qbomb)

The heat absorbed by the water can be calculated using the specific heat of water:

qwater=mcΔT

The heat absorbed by the calorimeter can be calculated from the heat capacity of the calorimeter:

qbomb=CΔT

Combine both equations into the first equation and substitute the known values, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute the terms of each multiplication and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)–(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)–(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)–50208 J+(420. J∘C×Tfinal)–8400 J]

Add the like terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, solve for Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

The answer should have three significant figures, so round to 22.7∘C.

8 0
2 years ago
H A and H B are both weak acids in water, and HA is a stronger acid than HB. Which of the following statements is correct? Selec
lubasha [3.4K]

Answer:

B is a stronger base than A^-, which is a stronger base than H2O, which is a stronger base than CI^-

Explanation:

The general equation for each acid is:

HA(aq) + H2O(ac) ⇄ H3O+(aq) + A-(aq)

HB(aq) + H2O(ac) ⇄ H3O+(aq) + B-(aq)

When these acids dissociate into its ions in water they lose a proton (H+), so they are proton donors (acids) and H2O is the proton acceptor (base). This reaction produces a conjugate acid and a conjugate base.

Conjugate base is what remains of the acid molecule after it loses a proton:

HA = acid         A- Conjugate base

HB = acid         B- Conjugate base

A conjugate acid is formed when the proton is transferred to the base

H2O = base                H3O+ = Conjugate acid

The stronger acid will produce a weaker base. According to this, if HA is a stronger acid than HB, A- would be the weaker base (B- is the stronger base).

Compared with water, A- and B- are stronger bases because when they compete for a proton they have much greater affinity for H+ than water does and the equilibrium position will lie far to the left. (HA and HB are weak acids)

Finally Cl- is the weakest base because it comes after dissociation of HCl which is a strong acid

HCl(aq) + H2O → H3O+(aq) + Cl-(aq)

Note there is no double arrows, equilibrium lies far to the right. A strong acid yields a weak conjugate base it means one that has a low affinity for a proton.

4 0
2 years ago
What mass of CaSO3 must have been present initially to produce 14.5 L of SO2 gas at a temperature of 12.5°C and a pressure of 1.
german
When the reaction equation is:

CaSO3(s) → CaO(s) + SO2(g)

we can see that the molar ratio between CaSO3 & SO2 is 1:1 so, we need to find first the moles SO2.

to get the moles of SO2 we are going to use the ideal gas equation:

PV = nRT

when P is the pressure =  1.1 atm

and V is the volume = 14.5 L 

n is the moles' number (which we need to calculate)

R ideal gas constant = 0.0821

and T is the temperature in Kelvin = 12.5 + 273 = 285.5 K

so, by substitution:

1.1 * 14.5 L = n * 0.0821 * 285.5

∴ n = 1.1 * 14.5 / (0.0821*285.5)

       = 0.68 moles SO2

∴ moles CaSO3 = 0.68 moles

so we can easily get the mass of CaSO3:

when mass = moles * molar mass

and we know that the molar mass of CaSO3= 40 + 32 + 16 * 3 = 120 g/mol


∴ mass = 0.68 moles* 120 g/mol = 81.6 g
7 0
2 years ago
How many grams are contained in a 0.893 mol sample of methane, ch4?
irina [24]

We are given with a compound, Methane (CH4), with a molar mass of 0.893 mol sample. We are tasked to solve for it's corresponding mass in g. We need to solve first the molecular weight of Methane, that is

C=12 g/mol

H=1g/mol

 

CH4= 12 g/mol +1(4) g/mol = 16 g/mol

With 0.893 mol sample, its corresponding mass is

g CH4= 0.893 mol x 16g/mol =14.288 g

Therefore, the mass of methane is 14.288 g

6 0
1 year ago
Bobby's mom sets a small pot of water on the stove and lights the burner. Ten minutes later, Bobby notices small bubbles and see
dolphi86 [110]
There are no answers, list them so I can answer. 
7 0
1 year ago
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