Salt and water is formed as product, when Arrhenius acid and Arrhenius base reacts and the reaction is known as neutralization reaction.
<h3>
What is Arrhenius acid-base reaction?</h3>
- The concept of acid and base based on the theory of ionization which was first proposed in 1884 by Svante Arrhenius
- According to Arrhenius, on dissociation in water the hydrogen-containing compounds which give H+ ions or protons are the acids and the hydroxide compounds which give OH− ions on dissociation in water are bases.
- This theory is applicable only to compounds which are dissolved in aqueous solution.
To learn more about Arrhenius acid-base reaction: brainly.com/question/15196401
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C and D are correct. This is the process of replication, and option E refers to repetition.
Answer:
a) 40,75 atm
b) 30,11 atm
Explanation:
The Ideal Gas Equation is an equation that describes the behavior of the ideal gases:
PV = nRT
where:
- P = pressure [atm]
- V = volume [L]
- n = number of mole of gas [n]
- R= gas constant = 0,08205 [atm.L/mol.°K]
- T=absolute temperature [°K]
<em>Note: We can express this values with other units, but we must ensure that the units used are the same as those used in the gas constant.</em>
The truncated virial equation of state, is an equation used to model the behavior of real gases. In this, unlike the ideal gas equation, other parameters of the gases are considered as the <u>intermolecular forces</u> and the <u>space occupied</u> by the gas
![\frac{Pv}{RT} = 1 + \frac{B}{v}](https://tex.z-dn.net/?f=%5Cfrac%7BPv%7D%7BRT%7D%20%3D%201%20%2B%20%5Cfrac%7BB%7D%7Bv%7D)
where:
- v is the molar volume [L/mol]
- B is the second virial coefficient [L/mol]
- P the pressure [atm]
- R the gas constant = 0,08205 [atm.L/mol.°K]
a) Ideal gas equation:
We convert our data to the adecuate units:
n = 5 moles
V = 3 dm3 = 3 L
T = 25°C = 298°K
We clear pressure of the idea gas equation and replace the data:
PV = nRT ..... P = nRT/V = 5 * 0,08205 * 298/3 =40,75 atm
b) Truncated virial equation:
We convert our data to the adecuate units:
n = 5 moles
V = 3 dm3 = 3 L
T = 25°C = 298°K
B = -156,7*10^-6 m3/mol = -156,7*10^-3 L/mol
We clear pressure of the idea gas equation and replace the data:
![\frac{Pv}{RT} = 1 + \frac{B}{v} ...... P = (1 + \frac{B}{v}) \frac{RT}{v}](https://tex.z-dn.net/?f=%5Cfrac%7BPv%7D%7BRT%7D%20%3D%201%20%2B%20%5Cfrac%7BB%7D%7Bv%7D%20......%20P%20%3D%20%281%20%2B%20%5Cfrac%7BB%7D%7Bv%7D%29%20%5Cfrac%7BRT%7D%7Bv%7D)
and v = 3 L/5 moles = 0,6 L/mol
![P = (1 + \frac{-156,7*10^{-3} }{0,6} ) \frac{0,08205*298}{0,6} = 30,11 atm](https://tex.z-dn.net/?f=P%20%3D%20%281%20%2B%20%5Cfrac%7B-156%2C7%2A10%5E%7B-3%7D%20%7D%7B0%2C6%7D%20%29%20%5Cfrac%7B0%2C08205%2A298%7D%7B0%2C6%7D%20%3D%2030%2C11%20atm)
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.
![400 mL \times \frac{0.996g}{1mL} =398g](https://tex.z-dn.net/?f=400%20mL%20%5Ctimes%20%5Cfrac%7B0.996g%7D%7B1mL%7D%20%3D398g)
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.
![100gH_2O \times \frac{550gGlucose}{398gH_2O} = 138 gGlucose](https://tex.z-dn.net/?f=100gH_2O%20%5Ctimes%20%5Cfrac%7B550gGlucose%7D%7B398gH_2O%7D%20%3D%20138%20gGlucose)
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.
Answer:
Coefficient of ![C_3H_8O_2=3](https://tex.z-dn.net/?f=C_3H_8O_2%3D3)
Coefficient of
=8
Explanation:
We are given that a reaction in which
reacts with ![KMnO_4](https://tex.z-dn.net/?f=KMnO_4)
We have to find the coefficient of each reactants in balanced reaction
![3C_3H_8O_2(aq)+8KMnO_4(aq)\rightarrow 3C_3H_2O_4K_2(aq)+8MnO_2(aq)+2KOH+8H_2O](https://tex.z-dn.net/?f=3C_3H_8O_2%28aq%29%2B8KMnO_4%28aq%29%5Crightarrow%203C_3H_2O_4K_2%28aq%29%2B8MnO_2%28aq%29%2B2KOH%2B8H_2O)
Coefficient is defined the constant value multiplied with a reactant in a reaction.
Coefficient of
=3
Coefficient of ![KMnO_4=8](https://tex.z-dn.net/?f=KMnO_4%3D8)
Coefficient of ![C_3H_2O_4K_2=3](https://tex.z-dn.net/?f=C_3H_2O_4K_2%3D3)
Coefficient of ![MnO_2=8](https://tex.z-dn.net/?f=MnO_2%3D8)
Coefficient of ![H_2O=8](https://tex.z-dn.net/?f=H_2O%3D8)
Coefficient of KOH=2
Hence, Coefficient of
and coefficient of ![KMnO_4=8](https://tex.z-dn.net/?f=KMnO_4%3D8)