Answer:
8CO + 17H2 = C8H18 + 8H2O
8, 17, 1, 8
0.24 moles of oxygen must be placed in a 3.00 L container to exert a pressure of 2.00 atm at 25.0°C.
The variables given are Pressure, volume and temperature.
Explanation:
Given:
P = 2 atm
V = 3 litres
T = 25 degrees or 298.15 K by using the formula 25 + 273.17 = K
R = 0.082057 L atm/ mole K
n (number of moles) = ?
The equation used is of Ideal Gas law:
PV = nRT
n = 
Putting the values given for oxygen gas in the Ideal gas equation, we get
n = 
= 0.24
Thus, from the calculation using Ideal Gas law it is found that 0.24 moles of oxygen must be placed in a container.
Ideal gas law equation is used as it tells the relation between temperature, pressure and volume of the gas.
Answer:
Explosive eruptions are favored by high gas content & high viscosity magmas (andesitic to rhyolitic magmas). The explosive bursting of bubbles fragments the magma into clots of liquid that cool as they fall through the air. These solid particles become pyroclasts or volcanic ash.
Answer:
The answer is "Option B".
Explanation:


![\to C \ CH_3COONa = \frac{(0.01\ mol + 5 \ E-4\ mol)}{(0.105\ L )}\\\\\to C \ CH_3COONa = 0.1 \ M\\\\\therefore Ka = ([H_3O^{+}]\times \frac{(0.1 + [H_3O^+]))}{(0.0905 - [H_3O^+])} = 1.75\ E-5\\\\\to 0.1[H_3O^+] + [H_3O^+]^2 = (1.75 E-5)\times (0.0905 - [H_3O^+])\\\\](https://tex.z-dn.net/?f=%5Cto%20C%20%5C%20CH_3COONa%20%3D%20%20%5Cfrac%7B%280.01%5C%20%20mol%20%2B%205%20%5C%20E-4%5C%20%20mol%29%7D%7B%280.105%5C%20L%20%29%7D%5C%5C%5C%5C%5Cto%20C%20%5C%20CH_3COONa%20%3D%200.1%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20Ka%20%3D%20%28%5BH_3O%5E%7B%2B%7D%5D%5Ctimes%20%5Cfrac%7B%280.1%20%2B%20%5BH_3O%5E%2B%5D%29%29%7D%7B%280.0905%20-%20%5BH_3O%5E%2B%5D%29%7D%20%3D%201.75%5C%20E-5%5C%5C%5C%5C%5Cto%200.1%5BH_3O%5E%2B%5D%20%2B%20%5BH_3O%5E%2B%5D%5E2%20%3D%20%281.75%20E-5%29%5Ctimes%20%280.0905%20-%20%5BH_3O%5E%2B%5D%29%5C%5C%5C%5C)
![\to [H_3O^+]^2 \ 0.1[H_3O^+] = 1.584\ E-6 - 1.75\ E-5[H_3O^+]\\\\\to [H_3O^+]^2 + 0.1000175[H_3O^+] - 1.584 \ E-6 = 0\\\\\to [H_3O^+] = 1.5835\ E-5 \ M\\\\\therefore pH = - \log [H_3O^+]\\\\\to pH = - \log (1.5835 \ E-5)\\\\ \to pH = 4.8004 > 4.7](https://tex.z-dn.net/?f=%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%5C%200.1%5BH_3O%5E%2B%5D%20%3D%201.584%5C%20%20E-6%20-%201.75%5C%20%20E-5%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%5BH_3O%5E%2B%5D%5E2%20%2B%200.1000175%5BH_3O%5E%2B%5D%20-%201.584%20%5C%20E-6%20%3D%200%5C%5C%5C%5C%5Cto%20%20%5BH_3O%5E%2B%5D%20%3D%201.5835%5C%20%20E-5%20%5C%20M%5C%5C%5C%5C%5Ctherefore%20pH%20%3D%20-%20%5Clog%20%5BH_3O%5E%2B%5D%5C%5C%5C%5C%5Cto%20%20pH%20%3D%20-%20%5Clog%20%281.5835%20%5C%20E-5%29%5C%5C%5C%5C%20%5Cto%20pH%20%3D%204.8004%20%3E%204.7)