A graph depicting a direct relation is a straight line and usually has positive slope. An inverse relation is a curve, typically concave up with negative slope.
A space-filling model shows the relative amount of space each atom takes up. In other words, a space-filling model can show relative sizes of atoms. However, unlike ball-and-stick or structural models, space-filling models do not show bond lengths clearly. Bonds are not really like sticks in a ball-and-stick model.
Volume of Cl₂(g) is produced at 1.0 atm and 540.°C=4.5×10^4 L
As per the evenly distributed response
2NaCl (g) ----> 2Na(l)+ Cl2(g)
Calculate the amount of Cl2 that was formed as indicated below:
Moles of Cl2 = 31.0 kg of Na x (1000* 1 * 1 / 1*23* 2)
= 673.9 mol
P is equal to 1.0 atm, and T is equal to 813.15 K
when converted to Kelvin by multiplying by a factor of 273.15.
Using Cl2 as an ideal gas, determine the in the following volume:
volume = nRT/P
= 673.9 * 0.0821 * 813.15/ 1
=4.5×10^4 L
As a result, the volume of Cl2 under the given circumstances =4.5×10^4 L
Learn more about Volume here:
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Answer:
Yes, it does, although only physically and not chemically.
Explanation:
If a volume of gas is way spread out, it won't collide with the other gas particles as often, reducing pressure and temperature because they lose kinetic energy to their surroundings when they don't collide.
If it is compressed, it increases temperature and pressure because the gas particles collide with each other and the walls of the container way more often than if they had more space.
Hope this answers your question.
P.S.
Fun fact, gas particles are actually moving at 300-400 meters per second at room temperature, they only slow down to walking speed at very low temperatures, like 10 Kelvin
Answer:
the pH of HCOOH solution is 2.33
Explanation:
The ionization equation for the given acid is written as:
Let's say the initial concentration of the acid is c and the change in concentration x.
Then, equilibrium concentration of acid = (c-x)
and the equilibrium concentration for each of the product would be x
Equilibrium expression for the above equation would be:
![\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}](https://tex.z-dn.net/?f=%5CKa%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BHCOO%5E-%5D%7D%7B%5BHCOOH%5D%7D)
From given info, equilibrium concentration of the acid is 0.12
So, (c-x) = 0.12
hence,
Let's solve this for x. Multiply both sides by 0.12
taking square root to both sides:
Now, we have got the concentration of ![[H^+] .](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20.)
![[H^+] = 0.00465 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%20%3D%200.00465%20M)
We know that, ![pH=-log[H^+]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%2B%5D)
pH = -log(0.00465)
pH = 2.33
Hence, the pH of HCOOH solution is 2.33.
