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MaRussiya [10]
2 years ago
5

El compuesto del cianuro

Chemistry
1 answer:
Evgen [1.6K]2 years ago
8 0

Answer:

CN -

Explanation:

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After your done with the second step,you will be left with a clear colorless liquid. What should be done to the liquid to finish
mafiozo [28]
In order to retrieve the soluble salt from the solution, crystallization must be carried out.<span />
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3 years ago
You have 1.88 moles of NaOH and an excess of H2SO4. How many grams of H2O will be produced?
KATRIN_1 [288]

Answer:

2 NaOH + H2SO4 2 H2O + Na2SO4

How many grams of sodium sulfate will be formed if you start with 200.0

grams of sodium hydroxide and you have an excess of sulfuric acid?

355.3 grams of Na2SO4

200.0 g NaOH 1 mol NaOH 1 mol Na2SO4 142.1 g Na2SO4

40.00 g NaOH 2 mol NaOH 1 mol Na2SO4

= 355.3 g

Explanation:

3 0
2 years ago
Provide only the major alkene product that results when n,n-dimethylhexan-2-amine undergoes cope elimination?
Fofino [41]

The major alkene  product that results when n,n-dimethylhexan-2-amine undergoes cope elimination is hexene or hex-1-ene.

The reaction in which an amine is oxidize to an intermediate called an N-oxide which , when heated , acts as base in an intramolecular elimination reaction. The oxidation of tertiary amine into N-oxide is called cope reaction.

This elimination gives the less substituted alkene along with more substituted alkene which is Zaitsev product.

Example: Cope elimination of  n,n-dimethylhexan-2-amine form hexene.

To learn more about alkene ,

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5 0
2 years ago
1 What is the total number of valence electrons in the Lewis structure of C104?
Lera25 [3.4K]
There are 32 valence electrons for the Lewis structure for ClO4

8 0
2 years ago
The vapor pressure of ethanol is 30°C at 98.5 mmHg and the heat of vaporization is 39.3 kJ/mol. Determine the normal boiling poi
Gelneren [198K]

Answer : The normal boiling point of ethanol will be, 348.67K or 75.67^oC

Explanation :

The Clausius- Clapeyron equation is :

\ln (\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})

where,

P_1 = vapor pressure of ethanol at 30^oC = 98.5 mmHg

P_2 = vapor pressure of ethanol at normal boiling point = 1 atm = 760 mmHg

T_1 = temperature of ethanol = 30^oC=273+30=303K

T_2 = normal boiling point of ethanol = ?

\Delta H_{vap} = heat of vaporization = 39.3 kJ/mole = 39300 J/mole

R = universal constant = 8.314 J/K.mole

Now put all the given values in the above formula, we get:

\ln (\frac{760mmHg}{98.5mmHg})=\frac{39300J/mole}{8.314J/K.mole}\times (\frac{1}{303K}-\frac{1}{T_2})

T_2=348.67K=348.67-273=75.67^oC

Hence, the normal boiling point of ethanol will be, 348.67K or 75.67^oC

3 0
3 years ago
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