Answer:
2.06 atm
Explanation:
The computation of the pressure inside the flask is shown below;
As per the ideal gas law
we know that
P V = nRT
where
V = 1.50L
T = 325 K
R = 0.0821
Now moles of N_2 is
= 3.25 ÷ 28
= 0.116 mol
Now
P = NRT ÷ V
= 0.116 × 0.0821 × 325 ÷ 1.50
= 2.06 atm
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.
Explanation:
Group 1A +1
Group 3A +3 or +1
Group 5A +5 or +3
The charges on an ion signifies the number of electrons that has been lost, gained or shared by an atom. These electrons are called valence electrons occupying the outermost shell of an atom.
Group 1A element will readily be willing to lose an electron giving them a charge of +1
Group 3A can lose either 1 or 3 electrons giving a charge of +3 or +1
Group 5A can lose either 5 or 3 electrons giving a charge of +5 or +3
The number of lost electron confers the stability on the configuration.
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Answer:
It's C
Explanation:
Most garnet found near Earth's surface forms when a sedimentary rock with a high aluminum content, such as shale, is subjected to heat and pressure intense enough to produce schist or gneiss.
