Answer:
Weather can have many classification, how hot how windy how cold how humid. There are however different classifications for different types of weather. Hope this helps :)
Answer:
A. 85.6 g
= 0.0856 kg.
B. 0.00027 mol/g
= 0.27 mol/kg.
C. 8.39 %
Explanation:
Given:
Molar concentration = 0.25 M
Molar weight of sucrose = 342.296 g/mol
Density of solution = 1.02 g/mL
Mass of water = 934.4 g.
Density in g/l = 1.020 g/ml * 1000ml/1 l
= 1020 g/l
Mass of solution in 1 l of solution = 1020 g
Mass of solution = mass of solvent + mass of solute
Mass of sucrose = 1020 - 934.4
= 85.6 g of sucrose in 1 l of solution.
A.
Density of sucrose = mass/volume
= molar mass/molar concentration
= 342.296 * 0.25
= 85.6 g/l
Number of moles = mass/molar mass
= 85.6/342.296
= 0.25 mol
B.
Molality = number of moles of solute/mass of solvent
= 0.25/934.4
= 0.00027 mol/g
C.
% mass of sucrose = mass of sucrose/total mass of solution * 100
= 85.6/1020 * 100
= 8.39 %
<u>Answer:</u> The specific heat of metal is 0.821 J/g°C
<u>Explanation:</u>
When metal is dipped in water, the amount of heat released by metal will be equal to the amount of heat absorbed by water.

The equation used to calculate heat released or absorbed follows:

......(1)
where,
q = heat absorbed or released
= mass of metal = 30 g
= mass of water = 100 g
= final temperature = 25°C
= initial temperature of metal = 110°C
= initial temperature of water = 20.0°C
= specific heat of metal = ?
= specific heat of water = 4.186 J/g°C
Putting values in equation 1, we get:
![30\times c_1\times (25-110)=-[100\times 4.186\times (25-20)]](https://tex.z-dn.net/?f=30%5Ctimes%20c_1%5Ctimes%20%2825-110%29%3D-%5B100%5Ctimes%204.186%5Ctimes%20%2825-20%29%5D)

Hence, the specific heat of metal is 0.821 J/g°C
Answer : A and D shows PROCESS of sweating
The given statement, some type of path is necessary to join both half-cells in order for electron flow to occur, is true.
Explanation:
Flow of electrons is possible with the help of a conducting medium like metal wire.
A laboratory device which helps in completion of oxidation and reduction-half reactions of a galvanic or voltaic cell is known as salt bridge. Basically, this salt bridge helps in the flow of electrons from anode to cathode and vice-versa.
If salt bridge is not present in an electrochemical cell, the electron neutrality will not be maintained and hence, flow of electrons will not take place.
Thus, we can conclude that the statement some type of path is necessary to join both half-cells in order for electron flow to occur, is true.