Answer:
The equilibrium concentration of hydrogen gas is 0.0010 M.
Explanation:
The equilibrium constant of the reaction =
}
Moles of hydrogen sulfide = 0.31 mol
Volume of the container = 4.1 L
![[concentration]=\frac{moles}{volume (L)}](https://tex.z-dn.net/?f=%5Bconcentration%5D%3D%5Cfrac%7Bmoles%7D%7Bvolume%20%28L%29%7D)
![[H_2S]=\frac{0.31 mol}{4.1 L}=0.076 M](https://tex.z-dn.net/?f=%5BH_2S%5D%3D%5Cfrac%7B0.31%20mol%7D%7B4.1%20L%7D%3D0.076%20M)

Initially
0.076 M
At equilibrium
(0.076-2x) 2x x
The expression of an equilibrium constant :
![K_c=\frac{[H_2]^2[S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BH_2%5D%5E2%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)

Solving for x:
x = 0.00051
The equilibrium concentration of hydrogen gas:
![[H_2]=2x=2\times 0.00051 M=0.0010 M](https://tex.z-dn.net/?f=%5BH_2%5D%3D2x%3D2%5Ctimes%200.00051%20M%3D0.0010%20M)
Answer:
4- A material that transfers heat energy more easily than another material will experience a greater rate of thermal energy loss than an object that does not transfer heat energy easily.
Explanation:
Thermal energy loss has to do with loss of heat energy by a body to another body or its environment. The aim of the process is usually the attainment of thermal equilibrium between the body and its environment.
On a cold day, a material that transfers thermal energy more easily will loose thermal energy faster than an object that does not transfer thermal energy. The rate of heat transfer of a body determines its rate of loss of thermal energy.
Answer: B.They are arranged in energy levels
Explanation: Electrons orbit around the nucleus of an atom, and are not spread around equally throughout an atom because there are a different amount of electrons that can fit in each energy level or electron shell (both are the same thing). For example the first energy level can hold up to 2 electrons, the second shell can hold up to 8 electrons, the third can hold up to 18, and so on represented by the equation 2(n^2)
Start studying Pressure - Volume Relationships in Gases (Boyle's Law). Learn vocabulary, terms ...Select<span> all that apply. V2 = k/P2 V2 = P1V1/P2 ... What </span>two variables<span> are </span>held constant when testing Boyle's Law in a manometer<span>? Temperature hope this helps
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Answer:
68.97g of milk must be added
Explanation:
To solve this problem we need to use the equation:
Q = C*m*ΔT -Coffee cup calorimeter equation-
This equation relates mass and change in temperature with heat of solution, Q.
The energy that decreases in the milk is the same that increases in the milk. That is:
Q(milk) = Q(tea)
Using the equation:
4.184J/molK* Mass milk * (65.0°C - 7.00°C) = 4.184J /molK * 200g (Tea) * (85.0°C - 65.0°C)
Mass milk * 58.0°C = 200g * 20.0°C
Mass milk = 4000g°C / 58.0°C
Mass milk =
68.97g of milk must be added