From the calculation, the concentration of the oxonium ions is 1 * 10^-6 M
<h3>What is concentration?</h3>
The term concentration refers to the amount of substances present in the solution. There are several units that can be used to show the concertation of a substances such as moles/liter, gram per liter, parts per billion, parts per billion, percentage etc.
Now we know that water is composed of the hydrogen and the hydroxide ions and the product of the both is generally known as the ion product of water and have a value of 1 * 10^-14.
If that be the case, we are in order to write the expression;
[H3O+] [OH-] = 1 * 10^-14
[H3O+] = 1 * 10^-14/ [OH-]
[H3O+] = 1 * 10^-14/ 1 x 10^ -8
[H3O+] = 1 * 10^-6 M
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Answer:
I would describe the jumping pattern of the green frog bellow as triangular or random it depend on you P.O.V.
Explanation:
Electrons are closer to the nucleus are in filled orbitals and are called core electrons. More energy which in nucleus called nuclear strOng energy to remove electron thars why its also a way harder too..
Answer:
<em>Molecules of different gases with the same mass and temperature always have the same average kinetic energy - E.</em>
Answer:
A = -213.09°C
B = 15014.85 °C
C = -268.37°C
Explanation:
Given data:
Initial volume of gas = 5.00 L
Initial temperature = 0°C (273 K)
Final volume = 1100 mL, 280 L, 87.5 mL
Final temperature = ?
Solution:
Formula:
The given problem will be solve through the Charles Law.
According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.
Mathematical expression:
V₁/T₁ = V₂/T₂
V₁ = Initial volume
T₁ = Initial temperature
V₂ = Final volume
T₂ = Final temperature
Conversion of mL into L.
Final volume = 1100 mL/1000 = 1.1 L
Final volume = 87.5 mL/1000 = 0.0875 L
Now we will put the values in formula.
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 1.1 L × 273 K / 5.00 L
T₂ = 300.3 L.K / 5.00 K
T₂ = 60.06 K
60.06 K - 273 = -213.09°C
2)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 280 L × 273 K / 5.00 L
T₂ = 76440 L.K / 5.00 K
T₂ = 15288 K
15288 K - 273 = 15014.85 °C
3)
V₁/T₁ = V₂/T₂
T₂ = V₂T₁ / V₁
T₂ = 0.0875 L × 273 K / 5.00 L
T₂ = 23.8875 L.K / 5.00 K
T₂ = 4.78 K
4.78 K - 273 = -268.37°C
