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allochka39001 [22]
3 years ago
15

1. Assume this experiment (after the extractions are complete) left you with 4 pure products (aspirin, acetaminophen and caffein

e, and the binder). Your lab mate distracted you after you had labeled the binder and now you can’t remember which of the remaining 3 products is which. What would be a simple test that you can perform in the laboratory to distinguish between the 3 remaining solids isolated from your experiments?
Chemistry
1 answer:
crimeas [40]3 years ago
4 0

Answer:

Explanation:

Out of aspirin, acetaminophen and caffeine, aspirin is an acid because it is acetyl saliciylic acid . Hence it can be tested with litmus paper .

acetaminophen contains phenolic functional group , hence it is a weak acid . It can be tested with any test with which phenol test are done,  like with neutral solution of ferric chloride .

caffeine is weak basic substance . It can also be tested with the help of testing a basic substance .

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The solid sugar crystals break apart in water as the sugar dissolves, but the individual sugar particles or molecules are still present and do not change as a result of dissolving in the water. The combined mass of the sugar and water shouldn't change.

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A tank holds 6.90 m3 of water. What is the volume in ft3? (3.28 ft = 1 m, Remember significant figures, but this answer does not
AleksAgata [21]

Volume of tank = 6.9 m^{3} (given)

Since, 1 m = 3.28 ft

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1 m^{3} = 35.287 ft^{3}

For 6.9 m^{3}:

6.9\times 35.287 ft^{3} = 243.4803 ft^{3}

The significant rule for multiplication, states that the number of significant figures in the answer obtained by multiplication is determined by the value with the lowest number of significant digits.

Since, the minimum number of decimal places in the above multiplication operation is 1 so, the final result must be upto 1 decimal place only.

243.4803 ft^{3} \simeq 243.5 ft^{3}

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Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

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3 years ago
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