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Valentin [98]
3 years ago
8

How many liters would be equal to 19 milliliters

Chemistry
1 answer:
zloy xaker [14]3 years ago
6 0
0.019 Hope this helps! :)
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Name the following compound: 2-ethyl-4-methylheptene 3,5-dimethyl-2-octene 2-ethyl-4-methylheptane 3-methyl-5-propyl-2-hexene
Nadusha1986 [10]

Answer:

3,5-dimethyl-2-octene

Explanation:

The parent chain will be choosen based on the highest value. In this case, if we count from top to bottom, we'll get seven carbon, however if we count from the second carbon, going left and then down, we'll get eight carbon. So the parent chain is octene

The double bond is located at the second carbon and the methyl groups are located on carbon 3 & 5. Since there are two methyl groups, we add di- in front of methyl to indicate two methyl groups present.

Note: The functional group has to be prioritise and it needed to be a part of the parent chain. In this case, the functional group is the double bond. (alkene)

5 0
4 years ago
WHAT IS THE STOCK NAME FOR FeSO^3
podryga [215]
Feso3 compound name

Iron(II) Sulfite FeSO3 Molecular Weight

Hope this helps!

Have a great day :)
4 0
3 years ago
5. What is the systemic name for the following structure?
topjm [15]

Answer:

You are not showing the question, but I believe the answer is cis-3,4-dimethyl-3-hexene.

Explanation:

since the substituents are on same side, it call cis. Followed by the name.

5 0
3 years ago
What is the Celsius temperature of 1 mole of a gas that has an average kinetic energy of 4,290 joules?
timofeeve [1]
Well the the answer is 70.8c but if you round it up it is 71c which I choice and got it correct so the answer is 71c

5 0
3 years ago
Salt in crude oil must be removed before the oil undergoes processing in a refinery. The
irina1246 [14]

Answer:

\large \boxed{0.64 \, \%}

Explanation:

Assume you are using 1 L of water.

Then you are washing 4 L of salty oil.

1. Calculate the mass of the salty oil

Assume the oil has a density of 0.86 g/mL.

\text{Mass of oil} = \text{4000 mL} \times \dfrac{\text{0.86 g}}{\text{1 mL}} = \text{3440 g}

2. Calculate the mass of salt in the salty oil

\text{Mass of salt} = \text{3440 g} \times \dfrac{\text{5 g salt}}{\text{100 g oil}} = \text{172 g salt}

3. Calculate the mass of salt in the spent water

\text{Mass of salt} = \text{1000 g water} \times \dfrac{\text{15 g salt}}{\text{100 g water}} = \text{150 g salt}

4. Mass of salt remaining in washed oil

Mass = 172 g - 150 g = 22 g  

5. Concentration of salt in washed oil

\text{Concentration} = \dfrac{\text{22 g}}{\text{3440 g}} \times 100 \, \% = \mathbf{0.64 \, \%}\\\\\text{The concentration of salt in the washed oil is $\large \boxed{\mathbf{0.64 \, \%}}$}

3 0
3 years ago
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