Answer:
The activation energy is
Explanation:
The gas phase reaction is as follows.
The rate law of the reaction is as follows.
The reaction is carried out first in the plug flow reactor with feed as pure reactant.
From the given,
Volume "V" =
Temperature "T" = 300 K
Volumetric flow rate of the reaction
Conversion of the reaction "X" = 0.8
The rate constant of the reaction can be calculate by the following formua.
Rearrange the formula is as follows.
The feed has Pure A, mole fraction of A in feed is 1.
= change in total number of moles per mole of A reacte.
Substitute the all given values in equation (1)
Therefore, the rate constant in case of the plug flow reacor at 300K is
The rate constant in case of the CSTR can be calculated by using the formula.
The feed has 50% A and 50% inerts.
Hence, the mole fraction of A in feed is 0.5
= change in total number of moles per mole of A reacted.
Substitute the all values in formula (2)
Therefore, the rate constant in case of CSTR comes out to be
The activation energy of the reaction can be calculated by using formula
In the above reaction rate constant at the two different temperatures.
Rearrange the above formula is as follows.
Substitute the all values.
Therefore, the activation energy is
Answer:
Explanation:
The number of molecules can be found by using the formula
N = n × L
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
N = 4.27 × 6.02 × 10²³
We have the final answer as
Hope this helps you
Answer:
There are 23 electrons in Fe3+
Answer:
Stoichiometric coefficient of hydrogen gas is 1.
Stoichiometric coefficient of palmitic acid is 1.
Explanation:
Addition of hydrogen to double bond is termed as hydrogenation reaction.
According to stoichiometry, 1 mole of palmitoleic acid reacts with 1 mole of hydrogen gas to give 1 mole of palmitic acid.
Stoichiometric coefficient of hydrogen gas is 1.
Stoichiometric coefficient of palmitic acid is 1.
The balanced chemical equation between HCl and is:
Moles of =
Moles of HCl required to neutralize :
Calculating the volume of HCl from moles and molarity:
Answer:- 117 mL of HCl are used.
Solution:- The balanced equation for the reaction of HCl with barium hydroxide is written as:
From above equation, HCl and react in 2:1 mol ratio.
We will calculate the moles of barium hydroxide on dividing its grams by its molar mass.
Molar mass of Barium hydroxide is given as 171.3 g per mol.
= 0.00584 mol
Using mol ratio we calculate the moles of HCl as:
= 0.01168 mol HCl
We know that molarity is moles of solute per liter of solution. We have 0.01168 moles of HCl and its molarity is 0.100 M. So, we can calculate the liters of HCl solution used on dividing the moles by molarity as and on multiplying by 1000 the liters are converted to mL since, 1 L = 1000 mL.
= 116.8 mL
It could be round to 117 mL.
So, 117 mL of HCl are required.
Answer:
Saturated solution
We should raise the temperature to increase the amount of glucose in the solution without adding more glucose.
Explanation:
Step 1: Calculate the mass of water
The density of water at 30°C is 0.996 g/mL. We use this data to calculate the mass corresponding to 400 mL.
Step 2: Calculate the mass of glucose per 100 g of water
550 g of glucose were added to 398 g of water. Let's calculate the mass of glucose per 100 g of water.
Step 3: Classify the solution
The solubility represents the maximum amount of solute that can be dissolved per 100 g of water. Since the solubility of glucose is 125 g Glucose/100 g of water and we attempt to dissolve 138 g of Glucose/100 g of water, some of the Glucose will not be dissolved. The solution will have the maximum amount of solute possible so it would be saturated. We could increase the amount of glucose in the solution by raising the temperature to increase the solubility of glucose in water.
The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated. If you want to increase the amount of glucose in the solution without adding more glucose, you can increase the temperature.
The solution made by adding 550 g of glucose to 400 mL of water at 30°C is saturated.
Since the solubility of glucose at 30°C is 125 g/100 g water, adding 550 g of glucose to 400 mL of water exceeds the maximum amount of glucose that can dissolve in the given amount of water.
To increase the amount of glucose in the solution without adding more glucose, you can increase the temperature. Higher temperatures generally increase the solubility of solutes in water. By increasing the temperature, you can dissolve more glucose in the solution.
Learn more about solubility here:
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Answer:
Lemon
HCI
Blood
Saliva
Bleach
NaOH
Explanation:
Blood 7.35-7.45
Bleach 12.6
Saliva 6.2-7.6
Lemon 2-3
HCI 3.01
NaOH 13