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.
Answer: Bb
Explanation:
The capital B (purbred black fur) is domaint the the lower case b(purebred white fur)
Reactant concentration, the physical state of the reactants, and surface area, temperature, and the presence of a catalyst are the four main factors that affect reaction rate.
Answer : The moles of methane gas could be, 
Solution :
According to the Graham's law, the rate of effusion of gas is inversely proportional to the square root of the molar mass of gas.
or,
![[\frac{(\frac{n_1}{t_1})}{(\frac{n_2}{t_2})}]=\sqrt{\frac{M_2}{M_1}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%28%5Cfrac%7Bn_1%7D%7Bt_1%7D%29%7D%7B%28%5Cfrac%7Bn_2%7D%7Bt_2%7D%29%7D%5D%3D%5Csqrt%7B%5Cfrac%7BM_2%7D%7BM_1%7D%7D)
where,
= rate of effusion of fluorine gas
= rate of effusion of methane gas
= moles of fluorine gas = 
= moles of methane gas = ?
= time = 12.3 min (as per question)
= molar mass of fluorine gas = 38 g/mole
= molar mass of methane gas = 16 g/mole
Now put all the given values in the above formula 1, we get:
![[\frac{(\frac{5.13\times 10^{-3}mol}{12.3min})}{(\frac{n_2}{12.3min})}]=\sqrt{\frac{16g/mole}{38g/mole}}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B%28%5Cfrac%7B5.13%5Ctimes%2010%5E%7B-3%7Dmol%7D%7B12.3min%7D%29%7D%7B%28%5Cfrac%7Bn_2%7D%7B12.3min%7D%29%7D%5D%3D%5Csqrt%7B%5Cfrac%7B16g%2Fmole%7D%7B38g%2Fmole%7D%7D)
Therefore, the moles of methane gas could be,
Answer:
Magma
Explanation:
Magma is part of molten rock at mid-ocean ridges.
Hope this helps!
