Heat required = 173 kJ
<h3>Further explanation</h3>
Given
56.0 g of ice
Temperatur at 263 K(-10 C) to 400 K(127 C)
Required
Heat needed
Solution
1. raise the temperature(-10 C to 0 C)⇒c ice=2.09 J/g C
2. phase change (ice to water)⇒Heat of fusion water=334 J/g
3. raise the temperature(0 C to 100 C)⇒c water= 4.18 J/g C
4. phase change(water to vapor)⇒heat of vaporization water=2260 J/g
5. raise the temperature(100 C to 127 C)⇒c vapor=2.09 J/g C
Total heat :
1170.4+18704+23408+126560+3160.08=173,002.48 J=173 kJ
<span>When water freezes, the molecules move farther apart. </span>
For which of the following activities might you want to hire a chemist?
Answer: D. testing a rock sample for gold content
Which of the following procedures involves a physical change in one of the substances?
Answer: C. separating a salt solution by evaporating the water
The equation Eºcell = 0.0592/n logK must be used to find n and also Eºcell
2 Al(s) + 3 Mg2+(aq) → 2 Al3+(aq) + 3 Mg(s) Al3+ +3e- --> Al Eº = -1.66 V Mg2+ +2e- -->Mg Eº = -2.37V
To balance the equation, 6 moles of electrons must be transferred (2 Al and 3 Mg). This will be the value of n in the equation.
To find Eºcell, you need the reduction potentials which should be given in a table, and given above. Eºcell = -1.66 - (-2.37) = 0.71 V log K = Eºcell x n/0.0592 = 0.71 x 6/0.0592 log K = 71.95 K = 10^71.95 K = 1.1x10^72
Answer:
Zr (Zirconium)
Explanation:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d2
