Answer:
600k
Explanation:
In this problem, we need to use a gas law that relates temperature to volume. The gas law to use here is the Charles’ law.
The Charles’ law posits that temperature and volume are directly proportional, provided that the pressure is kept constant.
Mathematically:
V1/T1 = V2/T2
We are looking at getting V2, hence we can write the mathematical equation as:
T2 = V2T1/V1
Asides the fact that we know that the gas is monoatomic, we do not know its volume. Let its initial volume be v. Since it expanded adiabatically, this means that its new volume is 2v
Hence: V1 = v , V2 = 2v , T1 = 300k and T2 is ?
Substituting these values, we have the following:
T2 = (2v * 300)/v
T2 = 600k
Explanation:
It is mainly the very low density of gases that make them bad conductors of heat. In liquids and solids atoms and molecules are densely packed and transfer of energy has much smaller distances to happen.
hope it helps u
Answer:
A polar covalent bond is a covalent bond in which the atoms have an unequal attraction for electrons and so the sharing is unequal. In a polar covalent bond, sometimes simply called a polar bond, the distribution of electrons around the molecule is no longer symmetrical.
Explanation:
Answer:
Ionic Exergonic
Explanation:
Exergonic reactions have negative Gibbs free energy and hence are spontaneous.
Endergonic reactions have negative Gibbs free energy and are non-spontaneous and low temperature.
Endergonic reactions require external supply of energy. But such reactions can be made spontaneous or favorable by coupling with exergonic reaction via a common intermediate.
Therefore, the correct answer is Ionic Exergonic.
Answer:
Explanation:
The balanced equation is
I₂(g) + Br₂(g) ⇌ 2IBr(g)
Data:
Kc = 8.50 × 10⁻³
n(IBr) = 0.0600 mol
V = 1.0 L
1. Calculate [IBr]
![\text{[IBr]} = \dfrac{\text{0.0600 mol}}{\text{1.0 L}} = \text{0.0600 mol/L}](https://tex.z-dn.net/?f=%5Ctext%7B%5BIBr%5D%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B0.0600%20mol%7D%7D%7B%5Ctext%7B1.0%20L%7D%7D%20%3D%20%5Ctext%7B0.0600%20mol%2FL%7D)
2. Set up an ICE table.
3. Calculate [I₂]
![\begin{array}{rcl}K_{\text{c}}&=&\dfrac{\text{[IBr]}^{2}} {\text{[I$_{2}$][Br]$_{2}$}}\\\\8.50 \times 10^{-2}&=&{\dfrac{(0.0600 - 2x)^{2}}{x^{2}}}& &\\\\0.2915x & = &{\dfrac{0.0600 - 2x}{x}}& &\\\\0.2915x & = &0.0600 - 2x\\\\2.2915x & = & 0.0600\\x & = & \textbf{0.026 18 mol/L}\\\end{array}\\](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brcl%7DK_%7B%5Ctext%7Bc%7D%7D%26%3D%26%5Cdfrac%7B%5Ctext%7B%5BIBr%5D%7D%5E%7B2%7D%7D%20%7B%5Ctext%7B%5BI%24_%7B2%7D%24%5D%5BBr%5D%24_%7B2%7D%24%7D%7D%5C%5C%5C%5C8.50%20%5Ctimes%2010%5E%7B-2%7D%26%3D%26%7B%5Cdfrac%7B%280.0600%20-%202x%29%5E%7B2%7D%7D%7Bx%5E%7B2%7D%7D%7D%26%20%26%5C%5C%5C%5C0.2915x%20%26%20%3D%20%26%7B%5Cdfrac%7B0.0600%20-%202x%7D%7Bx%7D%7D%26%20%26%5C%5C%5C%5C0.2915x%20%26%20%3D%20%260.0600%20-%202x%5C%5C%5C%5C2.2915x%20%26%20%3D%20%26%200.0600%5C%5Cx%20%26%20%3D%20%26%20%5Ctextbf%7B0.026%2018%20mol%2FL%7D%5C%5C%5Cend%7Barray%7D%5C%5C)
4. Convert the temperature to kelvins
T = (150 + 273.15) K = 423.15 K
5. Calculate p(I₂)
