Answer:
BaSO₄
Explanation:
It is possible to know if a bond is ionic or covalent using the electronegativity of the atoms in the bond. If electronegativity difference is higher than 1.8, the bond is ionic, if doesn't, bond is covalent.
CaI₂ has the Ca-I bond where electronegativity of Ca and I are 1 and 2.66. Difference of electronegativity is 1.66 → <em>Bond is covalent.</em>
COS has the C-O and C-S bonds where electronegativity of C, O and S are 2.55, 3.44 and 2.55. Difference of electronegativity are 0.89 and 0 → <em>Bonds are covalent.</em>
BaSO₄ has the Ba-O and O-S bonds where electronegativity of Ba, O and S are 0.89, 3.44 and 2.55. Difference of electronegativity are 2.55 and 0.89 → <em>Bonds are ionic and covalent respectively</em><em>.</em>
SF₆ has the S-F bond where electronegativity of S and F are 2.55 and 3.98. Difference of electronegativity is 1.43 → <em>Bond is covalent.</em>
Answer:
4.5 × 10^14 Hz
Explanation:
According to this question, chlorophyll pigment strongly absorbs light with a wavelength of 662nm. To calculate the frequency of this light, the following formula is used:
λ = v/f
Where;
λ = wavelength (m)
v = speed of light (3×10^8m/s)
f = frequency of light (Hz)
Since the λ = 662nm = 662 × 10^-9m,
f = v/λ
f = 3 × 10^8/662 × 10^-9
f = 0.004531 × 10^(8+9)
f = 0.0045 × 10^17
f = 4.5 × 10^14 Hz
Answer:
The major and minor products formed from the first structure have more alkyl groups on the C═C than those formed from the second structure.
The second structure has more hydrogens attached to the β carbons than the first structure.
Explanation:
It is possible to explain the reaction based on Zaitsev’s rule that states that an elimination will normally lead to the most stable alkene as the major product. This normally translates to it giving the most substituted alkene.
The two adjacent carbons in the first molecule are secondary. That means the two products will produce a disubstituted alkene but in the second molecule just one product will be disubstituted. Thus:
The first structure has more hydrogens attached to the β carbons than the second structure. <em>FALSE</em>. If the structure has more hydrogens will produce an alkene less stable.
The major and minor products formed from the first structure have more alkyl groups on the C═C than those formed from the second structure. <em>TRUE</em>. As the first structure have more alkyl groups the product is most stable.
The major and minor products formed from the second structure have more alkyl groups on the C═C than those formed from the first structure. <em>FALSE</em>. Is the opposite of the last option.
The second structure has more hydrogens attached to the β carbons than the first structure. <em>TRUE</em>. As the second structure has more hydrogens, the alkenes produced will be less substituted being less stable.
I hope it helps!
Answer:
I don't know the exact answer you are looking for
