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Rina8888 [55]
2 years ago
7

Benzaldehyde and benzyl alcohol can be distinguished by NMR. The proton on the aldehyde group will appear at approximately _____

ppm; the methylene peak on the alcohol is the only peak ______ ppm for either compound.
Chemistry
1 answer:
Tanya [424]2 years ago
8 0

Answer:

9-10 ppm.

0.2-0.4 ppm.

Explanation:

The proton on the aldehyde group will appear at approximately 9-10 ppm whereas the methylene peak on the alcohol is the only peak 0.2-0.4 ppm for either compound. Aldehydes and aromatics are quite distinctive in the Nuclear magnetic resonance (NMR). Aldehydes show up from 9-10 ppm, usually as a small singlet; aromatic protons show up from 6.5-8.5 ppm. NMR spectroscopy is the use of NMR to study the physical, chemical, and biological properties of matter.

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Consider the model of the nitrogen atom which electron configuration matches the model
san4es73 [151]

Answer:

           1s², 2s², 2p³  

Explanation:

The atomic number of Nitrogen is seven. So it contains seven protons and seven electrons in neutral form. Also, the electronic configuration cited above contains seven electrons among which two electrons are present in first shell and five electrons are present in valence shell respectively.

8 0
2 years ago
2. What effect does adding a neutron have on the atom’s mass?
anyanavicka [17]
It does<span>, however, change the </span>mass<span> of the nucleus. </span>Adding<span> or removing </span>neutrons<span>from the nucleus are how isotopes are created. Protons carry a positive electrical charge and they alone determine the charge of the nucleus.</span>
7 0
3 years ago
Acid indigestion is sometimes neutralized with an antacid such as magnesium hydroxide (Mg(OH)2). What products will be released
NemiM [27]

Answer:

Magnesium chloride and water  

Explanation:

Mg(OH)₂ + 2HCl ⟶                MgCl₂                +   2H₂O

                                    magnesium chloride        water

7 0
3 years ago
For each trial, compute the mol of titrant; (molarity x L) and keep the number of significant figures to 4.
MrMuchimi

Answer:

Trial     Number of moles

           

  1          0.001249mol

  2         0.001232mol

  3          0.001187 mol

Explanation:

To calculate the <em>number of moles of tritant</em> you need its<em> molarity</em>.

Since the<em> molarity</em> is not reported, I will use 0.1000M (four significant figures), which is used in other similar problems.

<em>Molarity</em> is the concentration of the solution in number of moles of solute per liter of solution.

In this case the solute is <em>NaOH</em>.

The formula is:

          Molarity=\dfrac{\text{Number of moles of solute}}{\text{Volume of solution in liters}}

Solve for the <em>number of moles:</em>

          \text{Number of moles}=Molarity\times Volume\text{ }in\text{ }liters

Then, using the molarity of 0.1000M and the volumes for each trial you can calculate the number of moles of tritant.

Trial    mL           liters          Number of moles

           

1          12.49       0.01249        0.01249liters × 0.1000M = 0.001249mol

2         12.32      0.01232         0.01232liters × 0.1000M = 0.001232mol

3          11.87       0.01187         0.01187liters × 0.1000M = 0.001187 mol

3 0
3 years ago
Read 2 more answers
Ammonia reacts with diatomic oxygen to form nitric oxide and water vapor: 4 NH3 + 5 O2 → 4 NO + 6 H2O When 40.0 g NH3 and 50.0 g
luda_lava [24]

Answer:

18.75 g of NH3.

Explanation:

The balanced equation for the reaction is given below:

4NH3 + 5O2 → 4NO + 6H2O

Next, we shall determine the masses of NH3 and O2 that reacted from the balanced equation.

This can be obtained as follow:

Molar mass of NH3 = 14 + (3x1) = 17 g/mol

Mass of NH3 from the balanced equation = 4 x 17 = 68 g

Molar mass of O2 = 16x2 = 32 g/mol

Mass of O2 from the balanced equation = 5 x 32 = 160 g

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Next, we shall determine the excess reactant. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, 40 g of NH3 will react with = (40 × 160)/68 = 94.12 g of O2.

From the calculations made above, we can see that it will take a higher amount of O2 i.e 94.12g than what was given i.e 50g to react completely with 40 g of NH3.

Therefore, O2 is the limiting reactant and NH3 is the excess reactant.

Next we shall determine the mass of excess reactant that reacted. This can be obtained as follow:

From the balanced equation above,

68 g if NH3 reacted with 160 g of O2.

Therefore, Xg of NH3 will react with 50 g of O2 i.e

Xg of NH3 = (68 × 50)/160

Xg of NH3 = 21.25 g

Therefore, 21.25 g of NH3 (excess reactant) were consumed in the reaction.

Finally, we shall determine mass of the remaining excess reactant as follow:

Mass of excess reactant = 40 g

Mass of excess reactant that reacted = 21.25 g

Mass of excess reactant remainig =?

Mass of excess reactant remainig = (Mass of excess reactant) – (Mass of excess reactant that reacted)

Mass of excess reactant remainig

= 40 – 21.25

= 18.75 g

Therefore, the mass of excess reactant remaining is 18.75 g of NH3.

8 0
3 years ago
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