Answer:
CH₅N
Explanation:
In the combustion, all of the C in the compound was used to produce CO₂ in a 1:1 ratio. Thus, the moles of CO₂ (MW 44.01 g/mol) produced equals the moles of C in the compound:
(44.0 g)(mol/44.01g) = 0.99977 mol CO₂ = 0.99977... mol C
Similarly, all of the H in the compound was used to produce H₂O in a ratio of 2H:1H₂O. The moles of H₂O (MW 18.02 g/mol) produced was:
(45.0 g)(mol/18.02g) = 2.497...mol H₂O
Moles of H is found using the molar ratio of 2H:1H₂O:
(2.497...mol H₂O)(2H/1H₂O) = 4.994...mol H
The ratio of H to C in the compound is:
(4.994...mol H)/(0.99977... mol C) = 5 H:C
Some NO₂ was produced from the N in the compound. Assuming a 1:1 ratio of C:N, the simplest empirical formula is: CH₅N.
I did this before and i had got an A no it but its been along time tho
Answer:
The temperature for ![\Delta G^o=0[/tex is [tex]T=328.6 K](https://tex.z-dn.net/?f=%5CDelta%20G%5Eo%3D0%5B%2Ftex%20is%20%5Btex%5DT%3D328.6%20K)
Explanation:
The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:
Where:
is Gibbs's energy in kJ
is the enthalpy in kJ
is the entropy in kJ/K
is the temperature in K
Solving:
For 
:
The correct answer for the question that is being presented above is this one:
Given that:
delta Tb = Kbm Kb H2O = 0.52 degrees C/m
<span>delta Tf = Kfm Kf H2O = 1.86 degrees C/m
</span>
We need to know the formula for Molality.
molality = mol solute / kg solvent
<span>We are given the amount of solute in grams
Since amount of solute is given in moles, we have to convert 25 g NaCl to moles. Divide by molar mass. </span>
<span>25 g NaCl / 58.5 g/mol = 0.427 mol </span>
<span>Then, use the formula for molality. </span>
<span>molality = mol solute / kg solvent </span>
<span>= 0.427 / 1 </span>
<span>= 0.427 m </span>
<span>Use now the formula to get the boiling point.</span>
<span>delta Tb = Kbm </span>
<span>= (0.52)(0.427) </span>
<span>= 0.22C </span>
Answer:
I think its b
Explanation:
but I wouldn't depend on this answer
