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NikAS [45]
3 years ago
11

Draw the two-carbon hydrocarbons propane, propene, and propyne. Classify each molecule, and explain the differences between them

.
Chemistry
1 answer:
Agata [3.3K]3 years ago
5 0
Two carbon compounds have the prefix eth-, the prefix pro- applies to three carbon atoms.
Propane: CH₃CH₂CH₃, alkane molecule, completely saturated and contains no double bonds
Propene: CH₃CH=CH₂, alkene molecule, single degree of unsaturation and contains one double bond
Propyne: CH₃C≡CH, alkyne molecule, two degrees of unsaturation and contains one triple bond
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Which of the following would increase friction?
KiRa [710]
B. Sand.

Sand would definitely increase friction because of the roughness of it surface. When its surface comes with another rough surface it would tend to increase friction. All other options given tend to reduce friction one way or the other.
5 0
2 years ago
Why can't liquids change volume but gasses can?
kupik [55]
Because liquids cant be condensed the way that gasses can for example in a tank of argon you can put 20 cubic feet because it can be be condensed but you could not fit 20 cubic feet of water because it can not be packed together . 
4 0
3 years ago
Read 2 more answers
In the Minnesota Department of Health set a health risk limit for acetone in groundwater of 60.0 μg/L . Suppose an analytical ch
siniylev [52]

Answer:

m = 4.7 μg

Explanation:

Given data:

density of acetone = 60.0  μg/L

Volume = 79.0 mL

Mass = ?

Solution:

Formula:

d = m/v

v = 79.0 mL × 1L /1000 mL

v = 0.079 L

Now we will put the values on formula:

d = m/v

60.0  μg/L = m/0.079 L

m = 60.0 μg/L × 0.079 L

m = 4.7 μg

So health risk limit for acetone = 4.7  μg

3 0
2 years ago
A alkaline earth metal is an element is groups 3-12 of the periodic table true or false?
Karo-lina-s [1.5K]

False Alkaline earth metals is group 2

5 0
2 years ago
What is the mass (grams) of salt in 10.0 m' of ocean water? ball park-4x10's (1.000 molsalt -58.44 g salt, 1.0 L ocean water -0.
koban [17]

Answer:

3.5 × 10⁵ g of salt

Explanation:

<em>What is the mass (grams) of salt in 10.0 m³ of ocean water?</em>

We have this data:

  • 1.000 mol salt is equal to 58.44 g salt
  • 1.0 L of ocean water contains 0.60 mol of salt

We will need the following relations:

  • 1 L = 1dm³
  • 1 m³ = 10³ dm³

We can use proportions:

10.0m^{3} .\frac{10^{3}dm^{3}  }{1m^{3} } .\frac{1L}{1dm^{3} } .\frac{0.60molSalt}{1.0L} .\frac{58.44gSalt}{1molSalt} =3.5 \times 10^{5} gSalt

8 0
3 years ago
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