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andrew-mc [135]
2 years ago
8

identify A and B, isomers of molecular formula C3H4Cl2, from the given 1H NMR data: Compound A exhibits peaks at 1.75 (doublet,

3 H, J = 6.9 Hz) and 5.89 (quartet, 1 H, J = 6.9 Hz) ppm. Compound B exhibits peaks at 4.16 (singlet, 2 H), 5.42 (doublet, 1 H, J = 1.9 Hz), and 5.59 (doublet, 1 H, J = 1.9 Hz) ppm.

Chemistry
1 answer:
siniylev [52]2 years ago
3 0

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem.

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an amorphous solid because the particles do not have a regular structure is the answer

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n the diagram shown, when an object ‘X’ is brought near the ring shaped magnet, the magnet moves away from it. Four friends are
kati45 [8]

Answer:

Akash

Explanation:

it could be a magnet with the same poles facing eachoher

6 0
2 years ago
An oxygen atom has a mass of and a glass of water has a mass of . Use this information to answer the question below. Be sure you
MariettaO [177]

Answer:

Number of moles of oxygen atoms having equal mass as  glass of water = 3.12 moles

<em>Note: The given question is missing some figures. Here is a complete similar question below.</em>

<em>An oxygen atom has a mass of 2.66*10^-23 g and a glass of water has a mass of 0.050 kg. What is the mass of 1 mole of oxygen atoms? Round your answer to 3 significant digits. How many moles of oxygen atoms have a mass equal to the mass of a glass of water? Round your answer to 2 significant digits.</em>

Explanation:

A mole of a substance contains the Avogadro number of particles = 6.02 * 10²³

Therefore a mole of oxygen atoms contains 6.02*10²³ atoms.

mass of 1 atom of oxygen = 2.66*10⁻²³ g

Mass of 1 mole of oxygen atoms = mass of 1 atom * number of atoms in 1 mole

Mass of 1 mole of oxygen atoms = 2.66*10⁻²³ g * 6.02*10²³  = 16. 01 g

Mass of a glass of water = 0.050 Kg or 50 g

To determine the number of moles of oxygen atoms that have a mass equal to a glass of water i.e. 50 g, the formula below is used;

<em>number of moles = mass/molar mass</em>

mass of oxygen atoms= 50 g, molar mass or mass of one mole of oxygen atoms = 16.01 g

Therefore, number of moles of oxygen atoms = 50 g / 16.01 g = 3.12 moles

4 0
2 years ago
The statement that percent yield can never be greater than theoretical yield is another example of the ________.
Gnom [1K]
We can rephrase the statement with a little more specificity in order to understand the answer here.

The mass of the products can never be more than the The mass that is expected.
3 0
2 years ago
1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl
kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0
2 years ago
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