A wet bandage is put on the forehead of a person suffering from high fever so that the temperature of the person slowly decreases from high to a little bit lower.
Pros it will sustain our daily needs and will never run out
<u>Answer:</u> 0.943 M of potassium hydroxide solution is required.
<u>Explanation:</u>
A neutralization reaction is defined as the reaction in which an acid reacts with a base to form a salt and water molecule.
To calculate the molarity of potassium hydroxide, the equation used is:
........(1)
where,
are the n-factor, molarity and volume of acid that is ![H_2SO_4](https://tex.z-dn.net/?f=H_2SO_4)
are the n-factor, molarity and volume of the base that is KOH
We are given:
![n_1=2\\M_1=0.506M\\V_1=23.3mL\\n_2=1\\M_2=?M\\V_2=25.0mL](https://tex.z-dn.net/?f=n_1%3D2%5C%5CM_1%3D0.506M%5C%5CV_1%3D23.3mL%5C%5Cn_2%3D1%5C%5CM_2%3D%3FM%5C%5CV_2%3D25.0mL)
Putting values in equation 1, we get:
![2\times 0.506\times 23.3=1\times M_2\times 25.0\\\\M_2=\frac{2\times 0.506\times 23.3}{1\times 25.0}\\\\M_2=0.943M](https://tex.z-dn.net/?f=2%5Ctimes%200.506%5Ctimes%2023.3%3D1%5Ctimes%20M_2%5Ctimes%2025.0%5C%5C%5C%5CM_2%3D%5Cfrac%7B2%5Ctimes%200.506%5Ctimes%2023.3%7D%7B1%5Ctimes%2025.0%7D%5C%5C%5C%5CM_2%3D0.943M)
Hence, 0.943 M of potassium hydroxide solution is required.
Answer:
See explanation.
Explanation:
Hello,
Here, considering the given flowrate, we perform the following units conversions for each case:
(i)
![11,900 \frac{L}{h} *\frac{0.264172 gal}{1L} *\frac{1h}{60min} =52.39\frac{gal}{min}](https://tex.z-dn.net/?f=11%2C900%20%5Cfrac%7BL%7D%7Bh%7D%20%2A%5Cfrac%7B0.264172%20gal%7D%7B1L%7D%20%2A%5Cfrac%7B1h%7D%7B60min%7D%20%3D52.39%5Cfrac%7Bgal%7D%7Bmin%7D)
(ii)
![11,900 \frac{L}{h} *\frac{1kg}{1L}*\frac{2.20462lbm}{1kg}=26,234\frac{lbm}{h}](https://tex.z-dn.net/?f=11%2C900%20%5Cfrac%7BL%7D%7Bh%7D%20%2A%5Cfrac%7B1kg%7D%7B1L%7D%2A%5Cfrac%7B2.20462lbm%7D%7B1kg%7D%3D26%2C234%5Cfrac%7Blbm%7D%7Bh%7D)
(iii)
![11,900 \frac{L}{h} *\frac{0.015gmolHCl}{1L}*\frac{1h}{60min}=2.975\frac{gmolHCl}{min}](https://tex.z-dn.net/?f=11%2C900%20%5Cfrac%7BL%7D%7Bh%7D%20%2A%5Cfrac%7B0.015gmolHCl%7D%7B1L%7D%2A%5Cfrac%7B1h%7D%7B60min%7D%3D2.975%5Cfrac%7BgmolHCl%7D%7Bmin%7D)
(iv) We find the kg of HCl per kg of acid solution:
![w_{HCl}=0.015\frac{gmolHCl}{Lsln}*\frac{36.45gHCl}{1gmolHCl} *\frac{1kgHCl}{1000gHCl}*\frac{1Lsln}{1kgsln} =5.4675x10^{-4}](https://tex.z-dn.net/?f=w_%7BHCl%7D%3D0.015%5Cfrac%7BgmolHCl%7D%7BLsln%7D%2A%5Cfrac%7B36.45gHCl%7D%7B1gmolHCl%7D%20%2A%5Cfrac%7B1kgHCl%7D%7B1000gHCl%7D%2A%5Cfrac%7B1Lsln%7D%7B1kgsln%7D%20%3D5.4675x10%5E%7B-4%7D)
(v)
![n_{HCl}=0.015\frac{gmolHCl}{L}*\frac{1000L}{1m^3} *1m^3=15gmolHCl](https://tex.z-dn.net/?f=n_%7BHCl%7D%3D0.015%5Cfrac%7BgmolHCl%7D%7BL%7D%2A%5Cfrac%7B1000L%7D%7B1m%5E3%7D%20%20%2A1m%5E3%3D15gmolHCl)
Best regards.