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3 years ago

Which statement correctly describes the actual yield and the theoretical yield of a reaction?

Chemistry

2 answers:

DochEvi [55]3 years ago

5 0

Answer:

Explanation:

edg 2020

Sunny_sXe [5.5K]3 years ago

4 0

Answer: <span>The actual yield depends on the reaction conditions, but the theoretical yield varies only with reactant amounts.

The actual yield has to be detemined from the real data.

The theoretical yield is calculated with the reactants amoounts and the theroretical molar ratios indicated by the chemical equation. Thus, the theoretical yield is the maximum possible yield, from the total reaction of the limiting reactant. The actual yield must be detemined under the particular contitions on which the reaction is performed.

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2 years ago

Find the enthalpy of neutralization of HCl and NaOH. 137 cm3 of 2.6 mol dm-3 hydrochloric acid was neutralized by 137 cm3 of 2.6

liraira [26]

Answer : The correct option is, (D) 89.39 KJ/mole

Explanation :

First we have to calculate the moles of HCl and NaOH.

$\text{Moles of HCl}=\text{Concentration of HCl}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

$\text{Moles of NaOH}=\text{Concentration of NaOH}\times \text{Volume of solution}=2.6mole/L\times 0.137L=0.3562mole$

The balanced chemical reaction will be,

$HCl+NaOH\rightarrow NaCl+H_2O$

From the balanced reaction we conclude that,

As, 1 mole of HCl neutralizes by 1 mole of NaOH

So, 0.3562 mole of HCl neutralizes by 0.3562 mole of NaOH

Thus, the number of neutralized moles = 0.3562 mole

Now we have to calculate the mass of water.

As we know that the density of water is 1 g/ml. So, the mass of water will be:

The volume of water = $137ml+137ml=274ml$

$\text{Mass of water}=\text{Density of water}\times \text{Volume of water}=1g/ml\times 274ml=274g$

Now we have to calculate the heat absorbed during the reaction.

$q=m\times c\times (T_{final}-T_{initial})$

where,

q = heat absorbed = ?

$c$ = specific heat of water = $4.18J/g^oC$

m = mass of water = 274 g

$T_{final}$ = final temperature of water = 325.8 K

$T_{initial}$ = initial temperature of metal = 298 K

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

$q=274g\times 4.18J/g^oC\times (325.8-298)K$

$q=31839.896J=31.84KJ$

Thus, the heat released during the neutralization = -31.84 KJ

Now we have to calculate the enthalpy of neutralization.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of neutralization = ?

q = heat released = -31.84 KJ

n = number of moles used in neutralization = 0.3562 mole

$\Delta H=\frac{-31.84KJ}{0.3562mole}=-89.39KJ/mole$

The negative sign indicate the heat released during the reaction.

Therefore, the enthalpy of neutralization is, 89.39 KJ/mole

3 0

3 years ago

a doctor's order is 0.125 g of ampicillin. the liquid suspension on hand contains 250 mg/5.0 ml. how many milliliters of the sus

disa [49]

0.125 g=(0.125 g)(1000 mg/1g)=125 mg.

Then, we need 125 mg of ampicillin.
5 ml of liquid suspension contains 250 mg of ampicilling , therefore:

5 ml----------------250 mg of ampicilling
x--------------------125 mg of ampicilling

x=(5 ml * 125 mg of ampicilling) / 250 mg of ampicilling=2.5 ml

Answer: we require 2.5 ml

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3 years ago

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