4 years ago

What is cis-3-Hexene?

1 answer:

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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An element's atomic number is 86. how many protons would an atom of this element have?

rusak2 [61]

The number of protons an element has is the same as the atomic number, so the element would have 86 protons.

8 0

3 years ago

How many grams of potassium sulphate are formed when 49g of sulphuric acid in solution are neutralised by potassium hydroxide?

Katena32 [7]

C - keep counting to the atoms

6 0

2 years ago

2Mg + O2à MgO For this unbalanced chemical equation, what is the coefficient for oxygen when the equation is balanced? A. 1 B. 2

Vikki [24]

Answer:

The answer to your question is the letter A. 1

Explanation:

Unbalanced chemical reaction

Mg + O₂ ⇒ MgO

Reactants Elements Products

1 Magnesium 1

2 Oxygen 1

Balanced chemical reaction

2Mg + O₂ ⇒ 2MgO

Reactants Elements Products

2 Magnesium 2

2 Oxygen 2

Conclusion

The coefficients of the balanced equation are 2, 1, 2

7 0

3 years ago

Read 2 more answers

10. When a piece of steel wool is burned it gains mass. Which explanation is consistent with the law of conservation of mass?

frez [133]

Answer:

Oh come on. Look at all - then look at A.

Explanation:

4 0

3 years ago

The vapor pressure of water at 65oC is 187.54 mmHg. What is the vapor pressure of a ethylene glycol (CH2(OH)CH2(OH)) solution ma

Pavlova-9 [17]

Answer:

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

Explanation:

Vapor pressure of water at 65 °C= $p_o= 187.54 mmHg$

Vapor pressure of the solution at 65 °C= $p_s$

The relative lowering of vapor pressure of solution in which non volatile solute is dissolved is equal to mole fraction of solute in the solution.

Mass of ethylene glycol = 22.37 g

Mass of water in a solution = 82.21 g

Moles of water= $n_1=\frac{82.21 g}{18 g/mol}=4.5672 mol$

Moles of ethylene glycol= $n_2=\frac{22.37 g}{62.07 g/mol}=0.3603 mol$

$\frac{p_o-p_s}{p_o}=\frac{n_2}{n_1+n_2}$

$\frac{187.54 mmHg-p_s}{187.54 mmHg}=\frac{0.3603 mol}{0.3603 mol+4.5672 mol}$

$p_s=173.83 mmHg$

173.83 mmHg is the vapor pressure of a ethylene glycol solution.

6 0

3 years ago