What is true of a solution of 1.0M HCL

Acid-catalyzed dehydration of an alcohol is an equilibrium situation. How was the reaction forced to completion in Acid-Catalyze

egoroff_w [7]

Answer:

In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.

Explanation:

In the acid-catalyzed dehydration of 2-methyl-2-butanol, the reaction can be driven to completion using Le Chatelier's principle. The reaction is driven to completion because the released water molecules form a strong bond with the acid used as a catalyst. As a result, the alkene produced can be distilled from the mixture.

3 0

3 years ago

Yeast converts glucose to ethanol and carbon dioxide during a process known as anaerobic fermentation. The chemical reaction is

Korolek [52]

Answer:

102g

Explanation:

To find the mass of ethanol formed, we first need to ensure that we have a balanced chemical equation. A balanced chemical equation is where the number of atoms of each element is the same on both sides of the equation (reactants and products). This is useful as only when a chemical equation is balanced, we can understand the relationship of the amount (moles) of reactant and products, or to put it simply, their relationship with one another.

In this case, the given equation is already balanced.

$\tex{C_6H_{12}O_6} \longrightarrow 2 \ C_2H_6O + 2 \ CO_2$

From the equation, the amount of ethanol produced is twice the amount of yeast present, or the same amount of carbon dioxide produced. Do note that amount refers to the number of moles here.

Mole= Mass ÷Mr

Mass= Mole ×Mr

Method 1: using the mass of glucose

Mr of glucose

= 6(12) +12(1) +6(16)

= 180

Moles of glucose reacted

= 200 ÷180

= $\frac{10}{9}$ mol

Amount of ethanol formed: moles of glucose reacted= 2: 1

Amount of ethanol

= $2(\frac{10}{9} )$

= $\frac{20}{9}$ mol

Mass of ethanol

= $\frac{20}{9} \times[2(12)+6+16]$

= $\frac{20}{9}(46)$

= 102 g (3 s.f.)

Method 2: using mass of carbon dioxide produced

Mole of carbon dioxide produced

= 97.7 ÷[12 +2(16)]

= 97.7 ÷44

= $\frac{977}{440}$ mol

Moles of ethanol: moles of carbon dioxide= 1: 1

Moles of ethanol formed= $\frac{977}{440}$ mol

Mass of ethanol formed

= $\frac{977}{440} \times[2(12)+6+16]$

= 102 g (3 s.f.)

Thus, 102 g of ethanol are formed.

Additional:

For a similar question on mass and mole ratio, do check out the following!

brainly.com/question/1685725

8 0

2 years ago

How many grams of carbonic acid were produced by the 3.00 g sample of NaHCO

stellarik [79]

I think the given is 3 g sample of NaHCO3. then if it will be reacted with an acid, it will produce H2CO3.
so the reaction NaHCO3 + HCl --> NaCl + H2CO3

mas of H2CO3 = 3 g NaHCO3 ( 1 mol NaHCO3 / 84 g ) ( 1 mol H2CO3 / 1 mol NaHCO3) ( 62.03 g / 1 mol )
mass of H2CO3 = 2.22 g H2CO3

7 0

3 years ago

Calculate the freezing point of a solution containing 5.0 grams of KCl and 550.0 grams of water. The molal-freezing-point-depres

yulyashka [42]

Answer: The freezing point of solution is -0.454°C

Explanation:

Depression in freezing point is defined as the difference in the freezing point of pure solution and freezing point of solution.

The equation used to calculate depression in freezing point follows:

$\Delta T_f=\text{Freezing point of pure solution}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\text{Freezing point of pure solution}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}$

where,

Freezing point of pure solution = 0°C

i = Vant hoff factor = 2

$K_f$ = molal freezing point elevation constant = 1.86°C/m

$m_{solute}$ = Given mass of solute (KCl) = 5.0 g

$M_{solute}$ = Molar mass of solute (KCl) = 74.55 g/mol

$W_{solvent}$ = Mass of solvent (water) = 550.0 g

Putting values in above equation, we get:

$0-\text{Freezing point of solution}=2\times 1.86^oC/m\times \frac{5\times 1000}{74.55g/mol\times 550}\\\\\text{Freezing point of solution}=-0.454^oC$

Hence, the freezing point of solution is -0.454°C

3 0

3 years ago

AB G C F G Y OBLAD

viktelen [127]

Blade resistance hair motion pull down

4 0

3 years ago