Answer:
B) 1-propanol has more hydrogen bonding than ethyl methyl ether.
Explanation:
1-propanol and ethyl methyl ether look very similar but have very different boiling points. They both have three carbon atoms but ethyl methyl ether has an oxygen atom.
Ethyl methyl ether has both London dispersion forces and dipole-dipole interactions.
1-Propanol has London dispersion forces, dipole-dipole interactions and hydrogen bonding.
1-propanol has more hydrogen bonding than ethyl methyl ether which explains why they both have different boiling points.
Answer:
ΔH° = + 6.2 kJ
Explanation:
Fe₂O₃(s) + 3CO(g) –––––> 2 Fe(s) + 3 CO₂(g); ΔH° = –26.8 kJ ......... ( 1 )
FeO(s) + CO(g) –––––> Fe(s) + CO₂(g); ΔH° = –16.5 kJ .........( 2 )
Multiplying equation ( 2 ) by 2 and subtracting it from ( 1 )
Fe₂O₃(s) + 3CO(g) -2 FeO(s) - 2CO(g) ––> 2 Fe(s) + 3 CO₂(g) - 2Fe(s) - 2CO₂(g) ΔH° = –26.8 kJ - ( 2 x –16.5 kJ )
Fe₂O₃(s) + CO(g) -2 FeO(s)––> CO₂(g) ΔH° = –26.8 kJ + 33 kJ
Fe₂O₃(s) + CO(g) ––>2 FeO(s) +CO₂(g) ΔH° = + 6.2 kJ
Required ΔH° = + 6.2 kJ Ans .
Answer:
A.) a proton is gained
Explanation:
A.) is correct. Protons are located in an atom's nucleus. As you travel to the right on a period, the elements are gaining more protons. For instance, magnesium, which is one space to the right of sodium, has one more proton than sodium.
B.) is incorrect. Like protons, as you travel to the right across a period, each element gains another electron. However, while the statement itself is true, <u>electrons are not located in an atom's nucleus</u>.
C.) is incorrect. Like protons, as you go across a period, the number of <u>neutrons in the nucleus increases</u>, not decreases.
D.) is incorrect. As noted in B.), the number of electrons in each element increases across the period. Therefore, as more electrons surround the nucleus, the <u>electron cloud size increases</u>.
<span>CH3OH(l) CH3OH(aq)
So the answer is A
</span>
Answer:
In the previous section, we discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, we were able to determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, we will explore how to apply these very same principles in order to derive the chemical formulas of unknown substances from experimental mass measurements.
Explanation:
tally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:
\displaystyle \%\text{H}=\frac{\text{mass H}}{\text{mass compound}}\times 100\%%H=
mass compound
mass H
×100%
\displaystyle \%\text{C}=\frac{\text{mass C}}{\text{mass compound}}\times 100\%%C=
mass compound
mass C
×100%
If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:
\displaystyle \%\text{H}=\frac{2.5\text{g H}}{10.0\text{g compound}}\times 100\%=25\%%H=
10.0g compound
2.5g H
×100%=25%
\displaystyle \%\text{C}=\frac{7.5\text{g C}}{10.0\text{g compound}}\times 100\%=75\%%C=
10.0g compound
7.5g C
×100%=75%
