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balandron [24]
2 years ago
12

Which of these prevents conduction from occurring?

Chemistry
2 answers:
cestrela7 [59]2 years ago
7 0
I’m pretty sure it’s C equal temperatures
uranmaximum [27]2 years ago
5 0

Answer:

I think it's c

Explanation:

if they were different temperatures then there would be conduction because other thermal energy will conduct to the lower temperature to reach a equal librium

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What is 2,4 hexadiene
Sunny_sXe [5.5K]
C6H10 Hope I helped! :P
8 0
3 years ago
Lithium diorganocopper (Gilman) reagents are prepared by treatment of an organolithium compound with copper(I) iodide. Decide wh
Ann [662]

Answer:

See explanation and image attached

Explanation:

The  Gilman reagent is a lithium and copper (diorganocopper) reagent with a general formula R2CuLi.  R is an alkyl or aryl group.

They are useful in the synthesis of alkanes because they react with organic halides to replace the halide group with an R group.

In this particular instance, we intend to synthesize propylcyclohexane. The structure of the  lithium diorganocopper (Gilman) reagent required is shown in the image attached to this answer.

7 0
3 years ago
How many grams of oxygen would be needed to completely react with 18 grams of methane (CH4)?
S_A_V [24]

Answer:

35.91 grams of oxygen

Explanation:

6 0
3 years ago
When 136g of glycine are dissolved in of a certain mystery liquid , the freezing point of the solution is lower than the freezin
loris [4]

The given question is incomplete. The complete question is:

When 136 g of glycine are dissolved in 950 g of a certain mystery liquid X, the freezing point of the solution is 8.2C lower than the freezing point of pure X. On the other hand, when 136 g of sodium chloride are dissolved in the same mass of X, the freezing point of the solution is 20.0C lower than the freezing point of pure X. Calculate the van't Hoff factor for sodium chloride in X.

Answer: The vant hoff factor for sodium chloride in X is 1.9

Explanation:

Depression in freezing point is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=T_f^0-T_f=8.2^0C = Depression in freezing point

K_f = freezing point constant

i = vant hoff factor = 1 ( for non electrolyte)

m= molality =\frac{136g\times 1000}{950g\times 75.07g/mol}=1.9

8.2^0C=1\times K_f\times 1.9

K_f=4.32^0C/m

Now Depression in freezing point for sodium chloride is given by:

\Delta T_f=i\times K_f\times m

\Delta T_f=20.0^0C = Depression in freezing point

K_f = freezing point constant  

m= molality = \frac{136g\times 1000}{950g\times 58.5g/mol}=2.45

20.0^0C=i\times 4.32^0C\times 2.45

i=1.9

Thus vant hoff factor for sodium chloride in X is 1.9

3 0
3 years ago
What is the concentration of a solution which is 2g in 150cm3 in g/dm3
Irina-Kira [14]

Answer:

13.33 g/dm³

Explanation:

Concentration (g/dm³)= mass(g) ÷ volume (dm³)

Now you need to convert 150 cm³ to dm³

1000cm³ = 1 dm³

thus, 150 cm3= 150 ÷ 1000

= 15dm³

and you already have mass in grams

so concentration  = 2 ÷ 0.15

= 13.33 g/dm³ and there you go.. solved ;)

8 0
3 years ago
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