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NeX [460]
4 years ago
12

What is the mass of a light bulb

Chemistry
1 answer:
DIA [1.3K]4 years ago
3 0
It is completely dependent on the type of light bulb you are talking about. There are many different kinds of light bulbs, I cant give you a solid answer.
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A graduated cylinder contains 17.5 ml of water. When a metal cube is placed onto the cylinder, it’s water level rises to 20.3 ml
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Answer:

20.3-17.5=2.8ml  1ml=1cm3   vol= 2.8cm3

Explanation:

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In fossil comparisons, it helps to know about similar:
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Fossils tell us when organisms lived, as well as provide evidence for the Fossils are the preserved remains or traces of animals, plants, and other organisms from the past. Fossils, by comparing the anatomies of both modern and extinct species, allows scientists to study the skin, hair, and organs of ancient creatures.

Explanation:

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Andrew drew this diagram to show a certain type of weather front. Which type of front is it?
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The masses of carbon and hydrogen in samples of four pure hydrocarbons are given above. The hydrocarbon in which sample has the
enot [183]

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

8 0
3 years ago
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