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3 years ago

6NaBr+1AlO3=3Na2O+2AlBr3 How many grams of NaBr would be needed in order to make 23.5 grams of AlBr3

Chemistry

1 answer:

Evgesh-ka [11]3 years ago

6 0

Answer:

23.5 grams of AlBr3 will be produced by 27.20 grams of NaBr

Explanation:

The balanced equation here is

6NaBr + 1AlO3 = 3Na2O + 2AlBr3

6 moles of NaBr are required to produce 2 moles of AlBr3

Mass of one mole of NaBr = 102.894 g/mol

Mass of one mole of AlBr3 = 266.69 g/mol

Mass of 6 moles of NaBr = 6*102.894 g/mol

Mass of two moles of AlBr3 = 2*266.69 g/mol

6*102.894 g NaBr produces 2*266.69 g of AlBr3

23.5 grams of AlBr3 will be produced by (6*102.894)/(2*266.69 )*23.5 = 27.20 grams of NaBr

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synthesize the following compound from benzene. use a diazonium salt as one of the synthetic intermediates.

irina [24]

Reacting nitrous acid with aromatic amines to produce diazonium salt is one of the most used processes. An aromatic amine called aniline reacts with nitrous acid to generate benzene diazonium chloride, a diazonium salt.

To create azo colors, benzene diazonium chloride combines with phenol, -napthol, aniline, and 2-methylaniline. These reactions result in the removal of an HCl molecule. Red organic dye is created when benzene and benzene diazonium chloride interact. Since the solutions created in this method do not last, the reactions involving a diazonium salt are always carried out with a freshly prepared solution. Diazonium salts are very flammable when they are solid because of their instability.

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7 0

1 year ago

This set contains the spectra of benzil, biphenyl, and bibenzyl, three compounds used in the melting point experiment. Using you

murzikaleks [220]

Answer:

Spectrum 1- Biphenyl

Spectrum 2 - Benzil

Spectrum 3 - Bibenzyl

Explanation:

For Benzil:

The Spectrum - 2 shows this compound.The C=O and C-H bond stretching observed at following values.

$C=O \,at\,1657cm^{-1}$

$C=H \,at\,3063cm^{-1}$

For Biphenyl:

Spectrum - 1 shows this compound.

There are stretching vibrations from $3061, 3032\,and \,1597 -1428cm^{1}$ due to C-H stretching and C-C stretching respectively of aromatic ring only.

For Bibenzyl:

Spectrum - 3 shows this compound

There are stretching vibrations at $3057,3026 and 1598 - 1449cm^{-1}$ due to C-H stretching and C-C stretching of aromatic ring.

Along with this, stretching vibrations at $2943 - 2917 cm^{-1}$ due to C-H of alkyl.

8 0

4 years ago

12.39 HBr can be added to alkenes in either the absence or presence of peroxides (producing either the Markovnikov or the anti-M

Alex777 [14]

Answer:

See explanation and image attached

Explanation:

The reaction of HBr with 1-butene is an addition reaction. The HBr adds across the double bond to yield a saturated halogenoalkane.

In the absence of peroxides, the reaction proceeds in accordance with Markovnikov rule which states that;''the negative part of the addendum, is attached to the carbon atom with the least number of hydrogen atoms attached.'' This occurs in the first reaction shown in the image.

In the presence of peroxides, the reaction proceeds in an anti-Markovnikov manner to yield the product shown in the second reaction.

4 0

3 years ago

Place the following in order of increasing standard molar entropy H2O(l) H2O(g) H2O(s) a)H2O(S)< H2O(l)

Black_prince [1.1K]

Answer : The order of increasing standard molar entropy is:

$H_2O(s)$

Explanation :

Entropy : It is defined as the measurement of randomness or disorderedness in a system.

The order of entropy will be,

As we are moving from solid state to liquid state to gaseous state, the entropy will be increases due to the increase in the disorderedness.

As we are moving from gaseous state to liquid state to solid state, the entropy will be decreases due to the decrease in the disorderedness.

From this we conclude that the solids have less disorder in the system than liquids because the molecules in the solid are closely or tightly packed as compared to liquid.

The liquids have less disorder in the system than gases because the molecules in the liquid are more close as compared to gases.

That means the more disorderedness in a system, the more will be the entropy.

So, the order of increasing standard molar entropy is:

$H_2O(s)$

7 0

3 years ago

The correct electron configuration for magnesium is: 1s 22s 22p 63s 3 True False

Dmitry [639]

Answer:

False

Explanation:

Magnesium is the element of second group and third period. The electronic configuration of magnesium is - 2, 8, 2 or $1s^22s^22p^63s^2$

There are 2 valence electrons of magnesium.

Only the valence electrons are shown by dots in the Lewis structure.

As, stated above, there are only two valence electrons of magnesium, so in the Lewis structure, two dots are made around the magnesium symbol.

Given that the electronic configuration is:- $1s^22s^22p^63s^3$ .

Orbital s cannot accommodate 3 electrons and also in magnesium it has $3s^2$ . Hence, the statement is false.

7 0

3 years ago