D, the rate increases as concentrations increase.
Typically, reaction rates decrease with time because reactant concentrations decrease as reactions are converted to products. Reaction rates generally increase when reactant concentrations are increased.
Parents SHOULDN'T avoid sharing their own mistakes with children.
Answer:
B. Conductivity can be used to calculate the salinity of the water.
I am 100% sure this is the correct answer :)
Answer:
Glycogen: It is made up of a large chain of molecules. ...
Cellulose: The cell wall of the plants is made up of cellulose. ...
Starch: It is formed by the condensation of amylose and amylopectin. ...
Insulin: It is made up of a number of fructofuranose molecules linked together in chains.
Explanation:
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