Chemical reaction: 4PBr₃(g) → P₄(g) + 6Br₂<span>(g).
</span>Pressure equilibrium constant (Kp) express the relationship between product pressures and reactant pressures. The partial pressures of gases are used to calculate pressure equilibrium constant.
Kp = (p(P₄) · p(Br₂)⁶) ÷ p(PBr₃)⁴.
p(P₄) - partial pressure of phosphorus.
p(Br₂) - partial pressure of bromine.
Answer:
1.2* 10³ rNe.
Explanation:
Given speed of neon=350 m/s
Un-certainity in speed= (0.01/100) *350 =0.035 m/s
As per heisenberg uncertainity principle
Δx*mΔv ≥\frac{h}{4\pi }
4π
h
..................(1)
mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg
6.22∗10
−23
20∗10
−3
=3.35∗10
−26
kg
substituating the values in eq. (1)
Δx =4.49*10^{-8}10
−8
m
In terms of rNe i.e 38 pm= 38*10^{-12}10
−12
Δx=\frac{4.49*10^{-8} }{38*10^{-12} }
38∗10
−12
4.49∗10
−8
=0.118*10^{4}10
4
* (rNe)
=1.18*10³ rN
= 1.2* 10³ rNe.
Explanation:
This is the answer
Answer:
the load is between the fulcrum and the handle
Explanation:
The answer 1.60217662 x 10^-19<span />
Answer:
(a) Ionic
(b) Nonpolar covalent
(c) Polar covalent
(d) Polar covalent
(e) Nonpolar covalent
(f) Polar covalent
<em>For those substances with polar covalent bonds, which has the least polar bond?</em> NO₂
<em>For those substances with polar covalent bonds, which has the most polar bond?</em> BF₃
Explanation:
<em>Are the bonds in each of the following substances ionic, nonpolar covalent, or polar covalent?</em>
The nature of a bond depends on the modulus of the difference of electronegativity (|ΔEN|) between the atoms that form it.
- If |ΔEN| = 0, the bond is nonpolar covalent.
- If 0 < |ΔEN| ≤ 2, the bond is polar covalent.
- If |ΔEN| > 2, the bond is ionic.
<em>(a) KCl</em> |ΔEN| = |EN(K) - EN(Cl)| = |0.8 - 3.0| = 2.2. The bond is ionic.
<em>(b) P₄</em> |ΔEN| = |EN(P) - EN(P)| = |2.1 - 2.1| = 0.0. The bond is nonpolar covalent.
<em>(c) BF₃</em> |ΔEN| = |EN(B) - EN(F)| = |2.0 - 4.0| = 2.0. The bond is polar covalent.
<em>(d) SO₂</em> |ΔEN| = |EN(S) - EN(O)| = |2.5 - 3.5| = 1.0. The bond is polar covalent.
<em>(e) Br₂</em> |ΔEN| = |EN(Br) - EN(Br)| = |2.8 - 2.8| = 0.0. The bond is nonpolar covalent.
<em>(f) NO₂</em> |ΔEN| = |EN(N) - EN(O)| = |3.0 - 3.5| = 0.5. The bond is polar covalent.
