Answer:
a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
Explanation:
It is given that a student performed an experiment where he dropped a nickel metal in to HCl solution. He observed the reaction and performed a splint test in the test tube that is filled with a gas which is formed while Nickle is dropped into the solution of HCl.
But the experiment that the student performed was incorrect. He must have observed the popping sound when the splint was placed in the test tube.
When the splint was added to the gas splint flared up. The hydrogen gas pops out when exposed to the flame.
Thus the correct option is (a).
ΔG⁰ = ΔH⁰ - T ΔS⁰
ΔG⁰ : Standard free energy of formation of acetylene
ΔH⁰ : Standard enthalpy of formation (226.7 kJ/mol)
ΔS⁰ : Standard entropy change (58.8 J / K. mol)
T : Temperature 25°C = 298 K (room temperature)
ΔG⁰ = 226.7 - (298 x 58.8 x 10⁻³) = 209.2 kJ /mol
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%
