pH=6.98
Explanation:
This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.
As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.
2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!
At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have
KW=[H3O+]⋅[OH−]=10−14
Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have
[H3O+]=√10−14=10−7M
The pH of pure water will thus be
pH=−log([H3O+])
pH=−log(10−7)=7
Now, let's assume that you're working with a 1.0-L solution of pure water and you add some 10
Using the mass/volume percentage method for percentages of the solution, you simply divide the grams of solute by the volume of the solution and multiply by 100 to get your percentage.
(75.0g/250mL)•100 = 30.0% solute
Answer:
1. n = 0.174mol
2. T= 26.8K
3. P = 1.02atm
4. V = 126.88L
Explanation:
1. P= 2.61atm
V = 1.69L
T = 36.1 °C = 36.1 + 273= 309.1K
R = 0.082atm.L/mol /K
n =?
n = PV / RT = (2.61x1.69)/(0.082x309.1)
n = 0.174mol
2. P = 302 kPa = 302000Pa
101325Pa = 1atm
302000Pa = 302000/101325 = 2.98atm
V = 2382 mL = 2.382L
T =?
n = 3.23 mol
R = 0.082atm.L/mol /K
T= PV /nR = (2.98x2.382)/(3.23x0.082) = 26.8K
3. P =?
V = 0.0250 m³ = 25L
T = 288K
n = 1.08mol
R = 0.082atm.L/mol /K
P = nRT/V = (1.08x0.082x288)/25 = 1.02atm
4. P = 782 torr
760Torr = 1 atm
782 torr = 782/760 = 1.03atm
V =?
T = 303K
n = 5.26 mol
R = 0.082atm.L/mol /K
V = nRT/P
V = (5.26x0.082x303)/1.03 = 126.88L
Molar mass NO₂ = 46.0 g/mol
1 mole -------- 46.0 g
2.0 moles ----- ?
Mass (NO₂) = 2.0 x 46.0 / 1
=> 92.0 g
hope this helps!
The North American plate is moving towards the west-southwest at about 2.3 centimeters every year mediated by the Mid-Atlantic Ridge, the spreading center, which gave rise to the Atlantic Ocean. The small Juan De Fuca plate, moving east-northeast at 4 centimeters every year, was once a component of much greater oceanic plates known as the Farallon plate.
The Farallon plate used to comprise what is now the Cocos plate of Mexico and Central America, and the Juan de Fuca plate in the region from N. Vancouver Island to the Cape Mendicino California, and a big sea floor tract in between. However, the middle portion of the Old Farallon plate disappeared underneath North America, it was subducted underneath California leaving the San Andreas fault system behind as the contact between the Pacific plates and North America.
The Juan De Fuca plate is still actively subducting underneath North America. Its movement is not smooth, however, rather sticky. The buildup of strain takes place until the fault dissociates and a few meters of Juan De Fuca get slid underneath North America in a big earthquake.
