Answer:
The correct answer is 1.21 L.
Explanation:
Based on the given information, the reaction will be,
CS2 (l) + 3Cl2 (g) ⇒ CCl4 (l) + S2Cl2 (l)
By using the standard values of the substances, the standard enthalpy of the reaction is,
ΔH° = [(-139.5) + (-58.5) – 0 – (87.3)] kJ/mol
= -285.3 kJ/mol
The amount of heat evolved for 3 moles of chlorine reacted us 285.3 kJ.
Now the number of moles of chlorine needed to react to produce 5.00 kJ is,
= 5.00 kJ × 3 mol Cl2/285.3 kJ
= 0.0526 mol Cl2
Now the volume of chlorine gas at 27degree C and 812 mmHg will be,
Volume = 0.0526 mol Cl2 × 0.0821 Latm/mol K × 300 K/ 1.07 atm
= 1.21 L
D. through measuring and observing !
Answer:
33.33 h
Explanation:
You know that
1 h = 60 min
If you divide both sides by 60 min, you get the conversion factor: 1 h/60 min = 1.
If you divide both sides by 1 h, you get the conversion factor: 1 = 60 min/1 h.
Both are conversion factors because they both equal one and multiplying a measurement by one does not change its value.
You choose the conversion factor that gives you the correct dimensions for your answer. It must have the correct dimensions on top (in the numerator),
Thus, to convert 2000 min to hours, you use the conversion factor with “h” on the top.
Answer:
18.8 g
Explanation:
The equation of the reaction is;
AgClO3(aq) + LiBr(aq)------>LiClO3(aq) + AgBr(s)
Number of moles of AgClO3 = 117.63 g/191.32 g/mol = 0.6 moles
Number of moles of LiBr = 10.23 g/86.845 g/mol = 0.1 moles
Since the molar ratio is 1:1, LiBr is the limiting reactant
Molar mass of solid AgBr = 187.77 g/mol
Mass of precipitate formed = 0.1 moles * 187.77 g/mol
Mass of precipitate formed = 18.8 g
Answer:
Figure 1 provides the rawest and most accurate data.
Explanation:
