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Lyrx [107]
3 years ago
10

How many grams of acetylene are produced by adding 3 moles of cac2

Chemistry
1 answer:
jolli1 [7]3 years ago
8 0

Answer:

78.12g of acetylene

Explanation:

Acetylene and calcium hydroxide are produced when H₂O reacts with CaC₂. The reaction is:

2 H₂O + CaC₂ → C₂H₂ + Ca(OH)₂

<em>Where 1 mole of C₂H₂ (acetylene) is produced per mole of CaC₂</em>

Thus, the addition of 3 moles of CaC₂ produces 3 moles of acetylene.

As molecular mass of acetylene is 26.04g/mol, grams of acetylene produced are:

3mol C₂H₂ × (26.04g / mol) = <em>78.12g of acetylene</em>

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8 0
3 years ago
Draw a Lewis structure for ketene, C2H2O , which has a carbon‑carbon double bond. Include all hydrogen atoms and show all unshar
Vladimir79 [104]

<u>Answer:</u> The lewis dot structure is attached below.

<u>Explanation:</u>

A Lewis dot structure is defined as the representation of atoms having electrons around the atom where electrons are represented as dots.

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The lewis dot structure of C_2H_2O is given in the image below.

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3 years ago
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7 0
4 years ago
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5 0
3 years ago
Can anyone help me on this please?
Bad White [126]
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one.  You can't
bluff through it.


7 0
4 years ago
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