Wind speed bc The faster the wind, the longer it blows, or the farther it can blow uninterrupted, the bigger the waves.
Answer:
53.29% of acetic acid is Oxygen
Explanation:
Step 1: Given data
Acetic acid it's molecular formula is HC2H3O2. This means it consists of 3 elements Carbon, Hydrogen and oxygen.
Step 2: the molar masses
The molecular mass of acetic acid is:
4* H = 4* 1.01 g/mole
2* C = 2*12 g/mole
2*O = 2* 16 g/mole
Total molar mass = 4+ 24+32 = 60.052 g/mole
Step 3: Calculate the mass percent
32 g of the 60.052 g is Oxygen
(32/60.052) *100% = 53.29%
53.29% of acetic acid is Oxygen.
Uhhh, well if the sun is out there is something called blackbody radiation. Wearing black clothes could absorb the sunlight just like solar panels on a house. ( black plates placed on houses) Plus you could get a heat stroke but its pretty rare.
Hey I can't see the questions properly
Data:
Molar Mass of HNO2
H = 1*1 = 1 amu
N = 1*14 = 14 amu
O = 3*16 = 48 amu
------------------------
Molar Mass of HNO2 = 1 + 14 + 48 = 63 g/mol
M (molarity) = 0.010 M (Mol/L)
Now, since the Molarity and ionization constant has been supplied, we will find the degree of ionization, let us see:
M (molarity) = 0.010 M (Mol/L)
Use: Ka (ionization constant) =
![5.0*10^{-4}](https://tex.z-dn.net/?f=5.0%2A10%5E%7B-4%7D%20)
![\alpha^2 (degree\:of\:ionization) = ?](https://tex.z-dn.net/?f=%20%5Calpha%5E2%20%28degree%5C%3Aof%5C%3Aionization%29%20%3D%20%3F)
![Ka = M * \alpha^2](https://tex.z-dn.net/?f=Ka%20%3D%20M%20%2A%20%5Calpha%5E2)
![5.0*10^{-4} = 0.010* \alpha^2](https://tex.z-dn.net/?f=5.0%2A10%5E%7B-4%7D%20%3D%200.010%2A%20%5Calpha%5E2%20)
![0.010\alpha^2 = 5.0*10^{-4}](https://tex.z-dn.net/?f=0.010%5Calpha%5E2%20%3D%205.0%2A10%5E%7B-4%7D)
![\alpha^2 = \frac{5.0*10^{-4}}{0.010}](https://tex.z-dn.net/?f=%5Calpha%5E2%20%3D%20%5Cfrac%7B5.0%2A10%5E%7B-4%7D%7D%7B0.010%7D%20)
![\alpha^2\approx500*10^{-4}](https://tex.z-dn.net/?f=%5Calpha%5E2%5Capprox500%2A10%5E%7B-4%7D)
![\alpha\approx\sqrt{500*10^{-4}}](https://tex.z-dn.net/?f=%5Calpha%5Capprox%5Csqrt%7B500%2A10%5E%7B-4%7D%7D%20)
![\alpha \approx 2.23*10^{-3}](https://tex.z-dn.net/?f=%5Calpha%20%5Capprox%202.23%2A10%5E%7B-3%7D)
Now, we will calculate the amount of Hydronium [H3O+] in nitrous acid (HNO2), multiply the acid molarity by the degree of ionization, we will have:
![[ H_{3} O^+] = M* \alpha](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%20M%2A%20%5Calpha%20)
![[ H_{3} O^+] = 0.010* 2.23*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%200.010%2A%202.23%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 0.0223*10^{-3}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%200.0223%2A10%5E%7B-3%7D)
![[ H_{3} O^+] \approx 2.23*10^{-5} \:mol/L](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%5Capprox%202.23%2A10%5E%7B-5%7D%20%5C%3Amol%2FL)
And finally, we will use the data found and put in the logarithmic equation of the PH, thus:
Data:
log10(2.23) ≈ 0.34
pH = ?
![[ H_{3} O^+] = 2.23*10^{-5}](https://tex.z-dn.net/?f=%5B%20H_%7B3%7D%20O%5E%2B%5D%20%3D%202.23%2A10%5E%7B-5%7D)
Formula:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)
Solving:
![pH = - log[H_{3} O^+]](https://tex.z-dn.net/?f=pH%20%3D%20-%20log%5BH_%7B3%7D%20O%5E%2B%5D)
![pH = -log2.23*10^{-5}](https://tex.z-dn.net/?f=pH%20%3D%20-log2.23%2A10%5E%7B-5%7D)
![pH = 5 - log2.23](https://tex.z-dn.net/?f=pH%20%3D%205%20-%20log2.23)
![pH = 5 - 0.34](https://tex.z-dn.net/?f=pH%20%3D%205%20-%200.34)
![\boxed{\boxed{pH = 4.66}}\end{array}}\qquad\quad\checkmark](https://tex.z-dn.net/?f=%5Cboxed%7B%5Cboxed%7BpH%20%3D%204.66%7D%7D%5Cend%7Barray%7D%7D%5Cqquad%5Cquad%5Ccheckmark)
Note:. The pH <7, then we have an acidic solution.