1) Answer is: molar mas of ammonia is 17.031 g/mol.
M(NH₃) = Ar(N) + 3 · Ar(H) · g/mol.
M(NH₃) = 14.007 + 3 · 1.008 · g/mol.
M(NH₃) = 17.031 g/mol.
2) Answer is: molar mas of lead(II) chloride is 278.106 g/mol.
M(PbCl₂) = Ar(Pb) + 2 · Ar(Cl) · g/mol.
M(PbCl₂) = 207.2 + 2 · 35.453 · g/mol.
M(PbCl₂) = 278.106 g/mol.
3) Answer is: molar mas of acetic acid is 60.052 g/mol.
M(CH₃COOH) = 2 · Ar(C) + 2 · Ar(O) + 4 · Ar(H) · g/mol.
M(CH₃COOH) = 2 · 12.0107 + 2 · 15.9994 + 4 · 1.008 · g/mol.
M(CH₃COOH) = 60.052 g/mol.
Answer:
The answer would be B. 1 joule/s.
Explanation:
The answer is B because the power in general is normally defined as energy over time. Watts are defined as <em><u>1 Watt = 1 Joule</u></em> per second (1W = 1 J/s) which means that 1 kW = 1000 J/s.
Another reason why it's B is because I had the same exact question in class, I took a screenshot of it a day ago:
<em><u>Hope this helps </u></em>
Use the General Gas Law
PV = nRT => n = PV/ RT
P= <span>10130.0 kPa
V= </span><span>50 L
R= </span><span>R = 8.314 L∙kPa/K∙mol
T= </span><span>300°C + 273 = 573 K
n = </span>10130.0 kPa 50 L / 8.314 L∙kPa/K∙mol <span>573 K
n = </span><span>106.32 mol</span>
Answer:
my bro what is the question first ?
Explanation:
Answer:
The process will be spontaneous above 702 K.
Explanation:
Step 1: Given data
- Standard enthalpy of the reaction (ΔH°): 308 kJ/mol
- Standard entropy of the reaction (ΔS°): 439 J/mol.K
Step 2: Calculate the temperature range in which the process will be spontaneous
The reaction will be spontaneous when the standard Gibbs free energy (ΔG°) is negative. We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T × ΔS°
When ΔG° < 0,
ΔH° - T × ΔS° < 0
ΔH° < T × ΔS°
T > ΔH°/ΔS°
T > (308,000 J/mol)/(439 J/mol.K)
T > 702 K
The process will be spontaneous above 702 K.