Answer:
14.3mL you require to reach the half-equivalence point
Explanation:
A strong acid as HClO₄ reacts with a weak base as CH₃CH₂NH₂, thus:
CH₃CH₂NH₂ + HClO₄ → CH₃CH₂NH₃⁺ + ClO₄⁻
As the reaction is 1:1, to reach the equivalence point you require to add the moles of HClO₄ equal to moles CH₃CH₂NH₂ you add originally. Also, half-equivalence point requires to add half-moles of CH₃CH₂NH₂ you add originally.
Initial moles of CH₃CH₂NH₂ are:
20.8mL = 0.0208L × (0.51mol CH₃CH₂NH₂ / 1L) =
0.0106moles CH₃CH₂NH₂
To reach the half-equivalence point you require:
0.0106moles ÷ 2 = 0.005304 moles HClO₄
As concentration of HClO₄ is 0.37M, volume you require to add 0.005304moles is:
0.005304 moles HClO₄ ₓ (1L / 0.37mol) = 0.0143L =
<h3> 14.3mL you require to reach the half-equivalence point</h3>
Answer:
P = 0.75 atm
Explanation:
Using Boyle's law
Given ,
<u>For Bulb A</u>
Pressure = 190 torr
The conversion of P(torr) to P(atm) is shown below:
So,
Pressure = 190 / 760 atm = 0.25 atm
Volume = 200 mL
For Bulb B
Pressure = 1 atm
Volume = 400 mL
For Bulb C
Volume = 1.00 L = 1000 mL
Pressure = 75.994 kPa
The expression for the conversion of pressure in KiloPascal to pressure in atm is shown below:
P (kPa) =
P (atm)
75.994 kPa =
atm
Pressure = 0.75 atm
Also, Total volume, V = 200 + 400 + 1000 mL = 1600 mL
Using above equation as:
We get, Total pressure, P = 0.75 atm
Answer:
![1.29*10^{24} mol](https://tex.z-dn.net/?f=1.29%2A10%5E%7B24%7D%20mol)
Explanation:
For this problem, you will have to convert moles to atoms. Use Avogadro's number.
![2.15 mol=\frac{6.022*10^{23} atoms}{1 mol}](https://tex.z-dn.net/?f=2.15%20mol%3D%5Cfrac%7B6.022%2A10%5E%7B23%7D%20atoms%7D%7B1%20mol%7D)
<u>Answer:</u> The temperature at which the food will cook is 219.14°C
<u>Explanation:</u>
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
![\frac{P_1}{T_1}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
![P_1=14.7psi\\T_1=273K\\P_2=(14.7+11.8)psi=26.5psi\\T_2=?](https://tex.z-dn.net/?f=P_1%3D14.7psi%5C%5CT_1%3D273K%5C%5CP_2%3D%2814.7%2B11.8%29psi%3D26.5psi%5C%5CT_2%3D%3F)
Putting values in above equation, we get:
![\frac{14.7psi}{273K}=\frac{26.5psi}{T_2}\\\\T_2=\frac{26.5\times 273}{14.7}=492.14K](https://tex.z-dn.net/?f=%5Cfrac%7B14.7psi%7D%7B273K%7D%3D%5Cfrac%7B26.5psi%7D%7BT_2%7D%5C%5C%5C%5CT_2%3D%5Cfrac%7B26.5%5Ctimes%20273%7D%7B14.7%7D%3D492.14K)
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
![T(K)=T(^oC)+273](https://tex.z-dn.net/?f=T%28K%29%3DT%28%5EoC%29%2B273)
![492.14=T(^oC)+273\\\\T(^oC)=219.14^oC](https://tex.z-dn.net/?f=492.14%3DT%28%5EoC%29%2B273%5C%5C%5C%5CT%28%5EoC%29%3D219.14%5EoC)
Hence, the temperature at which the food will cook is 219.14°C