Answer:
CnH2n-2
im pretty sure thats the answer
htdtshhthtdhthshAnswer:
yyrgtyht5gggggggggggggggggdggdd
Explanation:
KMnO4 has the incorrect set of oxidation numbers. Whenever there is an alkali metal, it has an oxidation number of +1. As you can see, K is said to have an oxidation number of +2, so it is incorrect.
Answer:- 4.36 kPa
Solution:- At constant volume, the pressure of the gas is directly proportional to the kelvin temperature.
![\frac{P_1}{T_1}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
Where the subscripts 1 and 2 are representing initial and final quantities.
From given data:
= 1.049 kPa
= ?
= 7.39 K
= 30.70 K
For final pressure, the equation could also be rearranged as:
![P_2=\frac{P_1T_2}{T_1}](https://tex.z-dn.net/?f=P_2%3D%5Cfrac%7BP_1T_2%7D%7BT_1%7D)
Let's plug in the values in it:
![P_2=\frac{1.049kPa(30.70K)}{7.39K}](https://tex.z-dn.net/?f=P_2%3D%5Cfrac%7B1.049kPa%2830.70K%29%7D%7B7.39K%7D)
= 4.36 kPa
So, the new pressure of the gas is 4.36 kPa.
Answer : The energy required is, 574.2055 KJ
Solution :
The conversions involved in this process are :
![(1):H_2O(s)(-10^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(95^oC)](https://tex.z-dn.net/?f=%281%29%3AH_2O%28s%29%28-10%5EoC%29%5Crightarrow%20H_2O%28s%29%280%5EoC%29%5C%5C%5C%5C%282%29%3AH_2O%28s%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%280%5EoC%29%5C%5C%5C%5C%283%29%3AH_2O%28l%29%280%5EoC%29%5Crightarrow%20H_2O%28l%29%2895%5EoC%29)
Now we have to calculate the enthalpy change or energy.
![\Delta H=[m\times c_{p,s}\times (T_{final}-T_{initial})]+n\times \Delta H_{fusion}+[m\times c_{p,l}\times (T_{final}-T_{initial})]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5Bm%5Ctimes%20c_%7Bp%2Cs%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D%2Bn%5Ctimes%20%5CDelta%20H_%7Bfusion%7D%2B%5Bm%5Ctimes%20c_%7Bp%2Cl%7D%5Ctimes%20%28T_%7Bfinal%7D-T_%7Binitial%7D%29%5D)
where,
= energy required = ?
m = mass of ice = 1 kg = 1000 g
= specific heat of solid water = ![2.09J/g^oC](https://tex.z-dn.net/?f=2.09J%2Fg%5EoC)
= specific heat of liquid water = ![4.18J/g^oC](https://tex.z-dn.net/?f=4.18J%2Fg%5EoC)
n = number of moles of ice = ![\frac{\text{Mass of ice}}{\text{Molar mass of ice}}=\frac{1000g}{18g/mole}=55.55mole](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%7BMass%20of%20ice%7D%7D%7B%5Ctext%7BMolar%20mass%20of%20ice%7D%7D%3D%5Cfrac%7B1000g%7D%7B18g%2Fmole%7D%3D55.55mole)
= enthalpy change for fusion = 6.01 KJ/mole = 6010 J/mole
Now put all the given values in the above expression, we get
![\Delta H=[1000g\times 4.18J/gK\times (0-(-10))^oC]+55.55mole\times 6010J/mole+[1000g\times 2.09J/gK\times (95-0)^oC]](https://tex.z-dn.net/?f=%5CDelta%20H%3D%5B1000g%5Ctimes%204.18J%2FgK%5Ctimes%20%280-%28-10%29%29%5EoC%5D%2B55.55mole%5Ctimes%206010J%2Fmole%2B%5B1000g%5Ctimes%202.09J%2FgK%5Ctimes%20%2895-0%29%5EoC%5D)
(1 KJ = 1000 J)
Therefore, the energy required is, 574.2055 KJ