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ale4655 [162]
4 years ago
15

What is cis-3-Hexene?

Chemistry
1 answer:
valentinak56 [21]4 years ago
7 0

cis-3-Hexene is a symmetrical cis-disubstituted alkene that can be prepared by the hydroboration of 3-hexyne followed by protonolysis

I got this from Sigmaaldrich.com.

Not my work.

Hope it helps

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A glass flask has a volume of 500 mL at a temperature of 20° C. The flask contains 492 mL of mercury at an equilibrium temperatu
OverLord2011 [107]

Answer:

101.63° C

Explanation:

Volume expansivity γa = γr -  γ g = 18 × 10⁻⁵ - 2.0 × 10⁻⁵ = 16 × 10⁻⁵ /K

v₂ - v₁ / v₁θ = 16 × 10⁻⁵ /K

(500 - 492 ) mL / (492 × 16 × 10⁻⁵) = θ

θ = 101.63° C

4 0
3 years ago
Three separate 3.5g blocks of al, cu, and fe at 25°c each absorb 0.505 kj of heat. which block reaches the highest temperature?
77julia77 [94]

The energy can be shown as:

Q = ms dT

Whereas, m is the mass of block

s is specific heat

dT is change in temperature.

Copper block having the lowest specific heat and thus having the higher change in temperature and therefore having the higher final temperature.

8 0
3 years ago
How many grams of water are needed to dissolve 27.8 g of ammonium nitrate NH4NO3 in order to prepare a 0.452 m solution?
Vanyuwa [196]

Answer: 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

Explanation:

Molality : It is defined as the number of moles of solute present per kg of solvent

Formula used :

Molality=\frac{n\times 1000}{W_s}

where,

n= moles of solute

Moles of NH_4NO_3=\frac{\text{Given mass}}{\text{Molar mass}}=\frac{27.8g}{80.0g/mol}=0.348moles  

W_s = weight of the solvent in g = ?

0.452=\frac{0.348\times 1000}{W_s}

W_s=770g

Thus 770 g water are needed to dissolve 27.8 g of ammonium nitrate NH_4NO_3 in order to prepare a 0.452 m solution

5 0
3 years ago
The following models represent three different mixtures of the same solid in water,
velikii [3]

Answer:

A

Explanation:

bcuz the water in it is little

3 0
3 years ago
WHAT IS THE PERCENT BY VOLUME OF ETHANOL IN A SOLUTION THAT CONTAINS 35 mL ETHANOL IN 115 mL OF WATER?
Dmitry [639]
We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.
3 0
3 years ago
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