Answer:
0.877 mol
Step-by-step explanation:
We can use the<em> Ideal Gas Law </em>to solve this problem.
pV = nRT Divide both sides by RT
n = (pV)/(RT)
Data:
p = 646 torr
V = 25.0 L
R = 0.082 06 L·atm·K⁻¹mol⁻¹
T = 22.0 °C
Calculations:
(a) <em>Convert the pressure to atmospheres
</em>
p = 646 torr × (1 atm/760 torr) = 0.8500 atm
(b) <em>Convert the temperature to kelvins
</em>
T = (22.0 + 273.15) K = 295.15 K
(c) <em>Calculate the number of moles
</em>
n = (0.8500 × 25.0)/(0.082 06 × 295.15)
= 0.877 mol
Answer : The value of equilibrium constant (K) is, 424.3
Explanation : Given,
Concentration of 
at equilibrium = 0.067 mol
Concentration of 
at equilibrium = 0.021 mol
Concentration of 
at equilibrium = 0.040 mol
The given chemical reaction is:
The expression for equilibrium constant is:
![K_c=\frac{[CH_3OH]}{[CO][H_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3OH%5D%7D%7B%5BCO%5D%5BH_2%5D%5E2%7D)
Now put all the given values in this expression, we get:
Thus, the value of equilibrium constant (K) is, 424.3
<u>Lithium Iodide</u><u>:</u>
~formed by the reaction of hydroxide with hydroiodic acid
Hope this helped you, have a good day bro cya)
Answer:
The isotope with the greatest number of protons is:
- <u>option D: Pu-239, with 94 protons</u>
Explanation:
The number of <em>protons</em> is the atomic number and is a unique number for each type of element.
You can tell the number of protons searching the element in a periodic table and reading its atomic number.
Thus, this is how you tell the number of protons or each isotope
Sample Chemical symbol Element atomic number # of protons
A Pa-238 Pa protactinium 91 91
B U-240 U uranium 92 92
C Np-238 Np neptunium 93 93
D Pu-239 Pu plutonium 94 94
I believe it is the Medial Temperal Lobe. It is part of the brain<span> known as the limbic system, which includes the hippocampus, the amygdala, the cingulate gyrus, the thalamus, the hypothalamus, the epithalamus, the mammillary body and other organs, many of which are of particular relevance to the processing of </span>memory<span>.</span>
