Answer:
0.50045
Explanation:
Because you have to look at a conversion table and move the decimal three places to the left. Hope this helps!
Answer:
*moles = molarity(M=moles/L) * volume (L)
so, you have 4.25 * 0.735 moles
*NH3
* M=n/v
*To determine the number of significant figures in a number use the following 3 rules: Non-zero digits are always significant. Any zeros between two significant digits are significant. A final zero or trailing zeros in the decimal portion ONLY are significant.
The presence of an atmosphere, temperature, and distance from the Sun do not affect a planet's gravity. Are the objects in the solar system still or are they in motion? The Sun's gravity pulls the planets in orbit around it, and some planets pull moons in orbit around them.
Answer:fH = - 3,255.7 kJ/mol
Explanation:Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
Answer: ![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. Pure solids are assumed to have a concentration of 1.
The given balanced equilibrium reaction is:

The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)
Thus the equilibrium constant expression for this reaction is ![K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCH_3Cl%5D%5Ctimes%20%5BOH%5E-%5D%7D%7B%5BCH_3OH%5D%5Ctimes%20%5BCl%5E-%5D%7D)