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2 years ago

Structural formula for 4-nonene and fluorine gas

Chemistry

2 answers:

goldenfox [79]2 years ago

6 0

I THINK it's <span>1,1-Difluorononane, or </span> $C_9H_{18}F_2$ .<span>
</span>

Savatey [412]2 years ago

6 0

I'm not sure how to draw it and show you, but I will type it out and hopefully you will understand.

4-nonene means the double bond is the fourth possible one:
CH 3 - CH 2 - CH 2 - CH=CH-CH 2 - CH 2 - CH 2 - CH 3
adding F2 means F2 are added across the double bond
-> CH 3 - CH 2 - CH 2 - CHF-CHF-CHF- CH 2 - CH 2 - CH 2 - CH 3
So: C9H18+F2 --> C9H18F2

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Please answer my questions. I really need them soon. I will post the links for other questions in the comments.

Assoli18 [71]

Explanation:

option option B is the correct answer of given statement helium-4(He)=2

6 0

2 years ago

How many grams of glucose are needed to prepare 144.3 mL of a 1.4%(m/v) glucose solution?

Montano1993 [528]

Answer:

2.0202 grams

Explanation:

1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.

so ?g glucose = 144.3 mL soln

Now apply the conversion factor, and you have:

?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).

so you have (144.3x1.4/100) g glucose= 2.0202 grams

6 0

3 years ago

Read 2 more answers

A student performs an experiment similar to what you will be doing in lab, except using titanium metal instead of magnesium meta

Genrish500 [490]

Answer:

n = 0.0022 mol

Explanation:

Moles is denoted by given mass divided by the molar mass ,

Hence ,

n = w / m

n = moles ,

w = given mass ,

m = molar mass .

From the information of the question ,

w = 0.108 g

As we known ,

The molar mass of titanium = 47.867 g / mol

The mole of titanium can be caused by using the above relation , i.e. ,

n = w / m

n = 0.108 g / 47.867 g / mol

n = 0.0022 mol

7 0

3 years ago

How many valence electrons are contained in N, O, and F, respectively? 5, 6, 7 7, 8, 9 3, 2, 1 14, 15, 18

trapecia [35]

Nitogen - 5
Oxygen - 6
Flourine - 7

7 0

3 years ago

Consider the reaction of ruthenium(III) iodide with carbon dioxide and silver. RuI3 (s) 5CO (g) 3Ag (s) Ru(CO)5 (s) 3AgI (s) Det

mixer [17]

Answer:

71.6 g of Ru(CO)₅ is the maximum mass that can be formed.

The limiting reactant is Ag

Explanation:

The reaction is:

RuI₃ (s) + 5CO (g) + 3Ag (s) → Ru(CO)₅ (s) + 3AgI (s)

Firstly we determine the moles of each reactant:

169 g . 1mol /481.77g = 0.351 moles of RuI₃

58g . 1mol /28g = 2.07 moles of CO

96.2g . 1mol/ 107.87g = 0.892 moles

Certainly, the excess reactant is CO, therefore, the limiting would be Ag or RuI₃.

3 moles of Ag react to 1 mol of RuI₃

Then 0.892 moles of Ag may react to (0.892 . 1) /3 = 0.297 moles

We have 0.351 moles of iodide and we need 0.297 moles, so this is an excess. In conclussion, Silver (Ag) is the limiting.

1 mol of RuI₃ react to 3 moles of Ag

Then, 0.351 moles of RuI₃ may react to (0.351 . 3) /1 = 1.053 moles

It's ok, because we do not have enough Ag. We only have 0.892 moles and we need 1.053.

5 moles of CO react to 3 moles of Ag

Then, 2.07 moles of CO may react to (2.07 . 3) /5 = 1.242 moles of Ag.

This calculate confirms the theory.

Now, we determine the maximum mass of Ru(CO)₅

3 moles of of Ag can produce 1 mol of Ru(CO)₅

Then 0.892 moles may produce (0.892 . 1) /3 = 0.297 moles

We convert moles to mass → 0.297 mol . 241.07g /mol = 71.6 g

8 0

3 years ago