The reaction has had a heat that is enthalpy of -22 kJ/mol. The exothermic process has been signaled by the negative sign.
The amount of energy that the system absorbs or releases to create the products is described as the heat of reaction.
The source of the reaction's heat is
H is equal to 3(413 Kj/mol) + 358 Kj/mol + 467 Kj/mol + 1070 Kj/mol = 3134 Kj/mol.
H prod equals 3(413 kj/mol) plus 347 kj/mol plus 358 kj/mol plus 467 kj/mol plus 745 kj/mol, or 3156 kj/mol.
H=3134 kj/mol - 3156 kj/mol = -22 Kj/mol
Negative findings point to an exothermic response.
A chemical process known as an exothermic reaction releases energy in the form of heat or light.
Learn more about exothermic reaction here-
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Answer:
1.12g/mol
Explanation:
The freezing point depression of a solvent for the addition of a solute follows the equation:
ΔT = Kf*m*i
<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>
<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>
<em>m is molality of the solution</em>
<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>
<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>
Replacing:
26.22°C = 5.35°Ckgmol⁻¹*m*1
4.90mol/kg = molality of the compound X
As the mass of the solvent is 100g = 0.100kg:
4.9mol/kg * 0.100kg = 0.490moles
There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:
0.551g / 0.490mol
= 1.12g/mol
<em>This result has no sense but is the result by using the freezing point of the solution = 13.4°C. Has more sense a value of -13.4°C.</em>
What is the solubility of barium chromate in parts per million?
*parts per million = Grams of Solute/grams of solution X 10^6 (which is ppm)
2.787 x 10^-3g/L x 1L/1000g x 10^6 = 0.02779, or 2.78 x 10^-2ppm
Answer in parts per million to three significant figures =2.78ppm
this is correct for the pearson mastering chemistry question
M(O₂)=20g
M(O₂)=32.0 g/mol
n(O₂)=20/32.0=0.625 mol
m(C)=12 g
M(C)=12.0 g/mol
n(C)=12/12.0=1.0 mol
2C + O₂ → 2CO
1 mol 0.625 mol 1 mol
0.625-0.5=0.125 mol
2CO + O₂ → 2CO₂
0.250 mol 0.125 mol 0.250 mol
n(CO)=1 mol - 0.250 mol = 0.750 mol
M(CO)=28.0 g/mol
m(CO)=0.750*28.0=21.0 g
n(CO₂)=0.250 mol
M(CO₂)=44.0 g/mol
m(CO₂)=0.250*44.0=11.0 g
Benzene at the same prssure ie the horizontal dotted line, benzene requires the minimal temperature hence its most volatile
