Answer:
The temperature change from the combustion of the glucose is 6.097°C.
Explanation:
Benzoic acid;
Enthaply of combustion of benzoic acid = 3,228 kJ/mol
Mass of benzoic acid = 0.570 g
Moles of benzoic acid = ![\frac{0.570 g}{122.12 g/mol}=0.004667 mol](https://tex.z-dn.net/?f=%5Cfrac%7B0.570%20g%7D%7B122.12%20g%2Fmol%7D%3D0.004667%20mol)
Energy released by 0.004667 moles of benzoic acid on combustion:
![Q=3,228 kJ/mol \times 0.004667 mol=15.0668 kJ=15,066.8 J](https://tex.z-dn.net/?f=Q%3D3%2C228%20kJ%2Fmol%20%5Ctimes%200.004667%20mol%3D15.0668%20kJ%3D15%2C066.8%20J)
Heat capacity of the calorimeter = C
Change in temperature of the calorimeter = ΔT = 2.053°C
![Q=C\times \Delta T](https://tex.z-dn.net/?f=Q%3DC%5Ctimes%20%5CDelta%20T)
![15,066.8 J=C\times 2.053^oC](https://tex.z-dn.net/?f=15%2C066.8%20J%3DC%5Ctimes%202.053%5EoC)
![C=7,338.92 J/^oC](https://tex.z-dn.net/?f=C%3D7%2C338.92%20J%2F%5EoC)
Glucose:
Enthaply of combustion of glucose= 2,780 kJ/mol.
Mass of glucose=2.900 g
Moles of glucose = ![\frac{2.900 g}{180.16 g/mol}=0.016097 mol](https://tex.z-dn.net/?f=%5Cfrac%7B2.900%20g%7D%7B180.16%20g%2Fmol%7D%3D0.016097%20mol)
Energy released by the 0.016097 moles of calorimeter combustion:
![Q'=2,780 kJ/mol \times 0.016097 mol=44.7491 kJ=44,749.1 J](https://tex.z-dn.net/?f=Q%27%3D2%2C780%20kJ%2Fmol%20%5Ctimes%200.016097%20mol%3D44.7491%20kJ%3D44%2C749.1%20J)
Heat capacity of the calorimeter = C (calculated above)
Change in temperature of the calorimeter on combustion of glucose = ΔT'
![Q'=C\times \Delta T'](https://tex.z-dn.net/?f=Q%27%3DC%5Ctimes%20%5CDelta%20T%27)
![44,749.1 J=7,338.92 J/^oC\times \Delta T'](https://tex.z-dn.net/?f=44%2C749.1%20J%3D7%2C338.92%20J%2F%5EoC%5Ctimes%20%5CDelta%20T%27)
![\Delta T'=6.097^oC](https://tex.z-dn.net/?f=%5CDelta%20T%27%3D6.097%5EoC)
The temperature change from the combustion of the glucose is 6.097°C.
The first major contributor to the Sun's apparent motion is the fact that Earth orbits the Sun while tilted on its axis. The Earth's axial tilt of approximately 23.5° ensures that observers at different locations will see the Sun reach higher-or-lower positions above the horizon throughout the year.
hope this helps ^^
Answer:
The gas with molar mass 83.9 g is Krypton, Kr
Explanation:
We know that as per Ghram's law of diffusion,
r'/ r= square root (M/M')
0r
Where
r= rate of diffusion of Cl2
r'= rate of diffusion of unknown gas= 0.920r
Also, M= molar mass of Cl2= 71 g/ mol
And, M'= molar mass of unknown gas which we have to find
Let us substitute the values in above equation
![\frac{0.920r}{r } =\sqrt{\frac{71}{M'} }](https://tex.z-dn.net/?f=%5Cfrac%7B0.920r%7D%7Br%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B71%7D%7BM%27%7D%20%7D)
solving we get M' = 83.9 g/mol
Therefore, the gas with molar mass 83.9 g is Krypton,Kr
Answer:
a)C2HBrClF3 = 197.35 g/mol
b)C12H14N2CL2 = 229.06g/mol
c)C8H10N4O2 = 194.22g/mol
d) CO(NH2)2=60.07 g/mol
e)C17H35CO2Na = 306.52 g/mol
Explanation:
Molar mass of a compound is equal to the sum of the atomic masses of the constituent elements.
a) C2HBrClF3
![Molar\ mass = 2(At. mass C)+1(at.mass H) +1(At. mass Br) + 1(At.mass Cl) + 3(At.mass F)\\=2(12.01 g/mol) + 1(1.01g/mol)+1(79.90 g/mol) +1(35.45g/mol)+3(18.99g/mol)=197.35g/mol](https://tex.z-dn.net/?f=Molar%5C%20mass%20%3D%202%28At.%20mass%20C%29%2B1%28at.mass%20H%29%20%2B1%28At.%20mass%20Br%29%20%2B%201%28At.mass%20Cl%29%20%2B%203%28At.mass%20F%29%5C%5C%3D2%2812.01%20g%2Fmol%29%20%2B%201%281.01g%2Fmol%29%2B1%2879.90%20g%2Fmol%29%20%2B1%2835.45g%2Fmol%29%2B3%2818.99g%2Fmol%29%3D197.35g%2Fmol)
b) C12H14N2CL2
![Molar\ mass = 12(C) + 14(H) + 2(N) + 2(Cl)\\\\=12(12.01) + 14(1.01) + 2(14.01) + 2(35.45) = 229.06g/mol](https://tex.z-dn.net/?f=Molar%5C%20%20mass%20%3D%2012%28C%29%20%2B%2014%28H%29%20%2B%202%28N%29%20%2B%202%28Cl%29%5C%5C%5C%5C%3D12%2812.01%29%20%2B%2014%281.01%29%20%2B%202%2814.01%29%20%2B%202%2835.45%29%20%3D%20229.06g%2Fmol)
c) C8H10N4O2
![Molar\ mass = 8(C) + 10(H) + 4(N) + 2(O)\\\\=8(12.01) + 10(1.01) + 4(14.01) + 2(16.00) =194.22g/mol](https://tex.z-dn.net/?f=Molar%5C%20%20mass%20%3D%208%28C%29%20%2B%2010%28H%29%20%2B%204%28N%29%20%2B%202%28O%29%5C%5C%5C%5C%3D8%2812.01%29%20%2B%2010%281.01%29%20%2B%204%2814.01%29%20%2B%202%2816.00%29%20%3D194.22g%2Fmol)
d) CO(NH2)2
![Molar\ mass = 1(C) + 1(O) + 2(N) + 4(H)\\\\=1(12.01) + 1(16.00) + 2(14.01)+4(1.01) =60.07 g/mol](https://tex.z-dn.net/?f=Molar%5C%20mass%20%3D%201%28C%29%20%2B%201%28O%29%20%2B%202%28N%29%20%2B%204%28H%29%5C%5C%5C%5C%3D1%2812.01%29%20%2B%201%2816.00%29%20%2B%202%2814.01%29%2B4%281.01%29%20%3D60.07%20g%2Fmol)
e) C17H35CO2Na
![Molar Mass = 18(C) + H(35) +2(O) + 1(Na)\\\\=18(12.01) + 35(1.01) + 2(16.00) + 1(22.99) =306.52 g/mol](https://tex.z-dn.net/?f=Molar%20Mass%20%3D%2018%28C%29%20%2B%20H%2835%29%20%2B2%28O%29%20%2B%201%28Na%29%5C%5C%5C%5C%3D18%2812.01%29%20%2B%2035%281.01%29%20%2B%202%2816.00%29%20%2B%201%2822.99%29%20%3D306.52%20g%2Fmol)