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3 years ago

Compare and contrast the functions of each of the following types of flasks: filter, Erlenmeyer, volumetric. (Site 2)

1 answer:

7 0

Filter flasks are also known as vacuum, suction or the Buchner flasks. They have thick walls and also have a short glass tube. The thick walls are designed to enable the filter withstand high pressures of vacuum applied to filter substances. Generally this is used for filtering.

While the Erlenmeyer flask also called as a conical flask, is a titration flask which consists of a conical body, a flat bottom, and round neck. This is used for used for general uses such as mixing, titrations, preparation of cultures, for recrystallization, and for supporting filter funnels.

<span>Lastly, the Volumetric flasks are graduated flasks which having markings for different volumes. They are calibrated accurately for a specific amount of liquid that can be contained in it hence this is specially used for storing precise amounts of liquid. </span>

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3 years ago

When water is electrolyzed, it splits into hydrogen and oxygen.

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The mass of water is equal to the combined mass of hydrogen and oxygen.

This is defined as the quantity of matter in a physical body. The electrolysis reaction of water can be seen below:

2 H₂O ---> 2 H₂ + O₂

We can deduce that 36 grams of H₂O dissociated to give 4 grams of H₂ and 32 grams of O₂ which option D was chosen as the appropriate choice.

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3 years ago

The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the

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Answer:

Sample B

Explanation:

In this case, we need to determine the empirical formula for each sample. The one that match the formula of the propene would be the sample.

Let's do Sample A:

C: 60 g; H: 12 g

1. Calculate moles:

We need the atomic weights of carbon (12 g/mol) and hydrogen (1 g/mol):

C: 60 / 12 = 5

H: 12 / 1 = 12

2. Determine number of atoms in the formula

In this case, we just divide the lowest moles obtained in the previous part, by all the moles:

C: 5 / 5 = 1

H: 12 / 5 = 2.4 or rounded to two

3. Write the empirical formula:

Now, the prior results, represent the number of atoms in the empirical formula for each element, so, we put them with the symbol and the atoms as subscripted:

C₁H₂ = CH₂

Therefore, sample A is not the same as propene.

Sample B:

C: 72 g H: 12 g

Following the same steps, let's determine the empirical formula for this sample

C: 72 / 12 = 6 ---> 6 / 6 = 1

H: 12 / 1 = 12 ----> 12 / 6 = 2

EF: CH₂

Sample C:

C: 84 g H: 10 g

C: 84 / 12 = 7 ----> 7 / 7 = 1

H: 10 / 1 = 10 ----> 10 / 7 = 1.4 or just 1

EF: CH

Sample D

C: 90 g H: 10 g

C: 90 / 12 = 7.5 -----> 7.5 / 7.5 = 1

H: 10 / 1 = 10 -------> 10 / 7.5 = 1.33 or just 1

EF: CH

Neither compound has the same empirical formula as C3H6, but C3H6 is a molecular formula, so, if we just simplify the formula we have:

C3H6 -----> CH₂

Therefore, sample B is the one that match completely. Sample B would be the one.

Hope this helps

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3 years ago