Answer:
The value of Kc is 660 (Option E) is correct
Explanation:
Step 1: Data given
Kp = 27
Temperature = 25.0 °C
Step 2: The balanced equation
2 NO(g) + Br2(g) ⇄ 2 NOBr(g)
Step 3: Calculate Kc
Kp = Kc * (RT)^Δn
⇒ with Kp = 27
⇒ with Kc = TO BE DETERMINED
⇒ with R = the gas constant = 0.08206 L*atm/mol*K
⇒ with T = The temperature = 25 °C = 298 K
⇒ Δn = the difference in moles = -1
27 = Kc * (0.08206*298)^-1
Kc = 660
The value of Kc is 660 (Option E) is correct
Answer:
= 5.26 × 10^-7 M
Explanation:
We know that;
[H3O+][OH-] = 1 x 10^-14
Therefore;
Given; OH− concentration = 1.9x10^−8 M
Then; [H3O+] = (1 x 10^-14)/[OH-]
= (1 x 10^-14)/(1.9x10^−8)
= 5.26 × 10^-7 M
Answer:
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
Explanation:
Atomic radii trend along group:
As we move down the group atomic radii increased with increase of atomic number. The addition of electron in next level cause the atomic radii to increased. The hold of nucleus on valance shell become weaker because of shielding of electrons thus size of atom increased.
Consider the example of sodium and potassium.
Sodium is present above the potassium with in same group i.e, group one.
The atomic number of sodium is 11 and potassium 19.
So potassium will have larger atomic radius as compared to sodium.
Atomic radius of sodium = 227 pm
Atomic radius of potassium = 280 pm
C. Homogeneous mixtures These alloys are homogeneous mixtures because they have a uniform composition throughout.
Answer:
11 atoms
Explanation:
The formula for lead(II) bicarbonate is: Pb(HCO3)2.
Atoms in Pb(HCO3)2:
Pb=1, H=1×2, C=1×2, O=3×2
1+1×2+1×2+3×2
=1+2+2+6
=11 atoms