The final temperature = 36 °C
<h3>Further explanation</h3>
The balanced combustion reaction for C₆H₆
2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ
MW C₆H₆ : 78.11 g/mol
mol C₆H₆ :
![\tt \dfrac{8.5}{78.11}=0.109](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B8.5%7D%7B78.11%7D%3D0.109)
Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol
![\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol](https://tex.z-dn.net/?f=%5Ctt%20%5Cdfrac%7B0.109%7D%7B2%7D%5Ctimes%206542%3D356.539~kJ%2Fmol)
Heat transferred to water :
Q=m.c.ΔT
![\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC](https://tex.z-dn.net/?f=%5Ctt%20356.539%3D5.691~kg%5Ctimes%204.18~kj%2Fkg%5EoC%5Ctimes%20%28t_2-21%29%5C%5C%5C%5Ct_2-21%3D15%5Crightarrow%20t_2%3D36%5EoC)
The human gametes differ from all the other cells in the body because the somatic cells have a complete set of chromosomes
Answer:
0.595 M
Explanation:
The number of moles of water in 1L = 1000g/18g/mol = 55.6 moles of water.
Mole fraction = number of moles of KNO3/number of moles of KNO3 + number of moles of water
0.0194 = x/x + 55.6
0.0194(x + 55.6) = x
0.0194x + 1.08 = x
x - 0.0194x = 1.08
0.9806x= 1.08
x= 1.08/0.9806
x= 1.1 moles of KNO3
Mole fraction of water= 55.6/1.1 + 55.6 = 0.981
If
xA= mole fraction of solvent
xB= mole fraction of solute
nA= number of moles of solvent
nB = number of moles of solute
MA= molar mass of solvent
MB = molar mass of solute
d= density of solution
Molarity = xBd × 1000/xAMA ×xBMB
Molarity= 0.0194 × 1.0627 × 1000/0.981 × 18 × 0.0194×101
Molarity= 20.6/34.6
Molarity of KNO3= 0.595 M
Answer:
Its pH value increases.
Explanation:
pH is the measure of alkalinity or acidity of a compound.
pH = - log [H+]
and pH + pOH = 14
where pOH is the measure of basicity of a solution, given by -log[OH-]
As a solution gets more basic that is higher [OH-], the pH increases, and on the other hand, as the pH of a solution decreases by one pH unit, the concentration of H+ increases by ten times.
V1 = 2.0 L
T1 = 25.0 oC = 298 K V2 = V1T2 = (2.0 L)(244 K) = 1.6 L
V2 = ? t1(298 K)
T2 = –28.9 oC = 244 K