Answer:
Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
Explanation:
In writing the cell notation for an electrochemical cell, the anode is written on the left hand side while the cathode is written on the right hand side. The two half cells are separated by two thick lines which represents the salt bridge.
For the cell discussed in the question; the Mn(s)/Mn^2+(aq) is the anode while the Co^2+(aq)/Co(s) half cell is the cathode.
Hence I can write; Mn(s)/Mn^2+(aq)//Co^2+(aq)/Co(s)
In order to balance this equation you need to count each element and how many of the individual elements are in the equation.
_H2+N2=2 NH3
You multiply the 2 (Which is the coefficient) by the 3 (which is the subscript) This would equal 6 which indicated there are 6 hydrogen atoms on the right side so the left side should also have 6 hydrogen atoms
The missing coefficient on the left side must multiple the 2 to become 6 hydrogen
Answer=3
Answer:
A solvent is a substance which dissolves a solute. When a solvent dissolves into a solute, it creates a solution
The enthalpy change for melting ice is called the entlaphy of fusion. Its value is 6.02 kj/mol. This means for every mole of ice we melt we must apply 6.02 kj of heat. We can calculate the heat needed with the following equation:
Q = N x ΔH
where:
Q = heat
N = moles
ΔH = enthalpy
In this problem we would like to calculate the heat needed to melt 35 grams of ice at 0 °C. This problem can be broken into three steps:
1. Calculate moles of water
2. multiply by the enthalpy of fusion
3. Convert kJ to J.
Step 1 : Calculate moles of water
![[ 75g ] x (\frac{1 mol}{18.02g} ) =](https://tex.z-dn.net/?f=%5B%2075g%20%5D%20x%20%28%5Cfrac%7B1%20mol%7D%7B18.02g%7D%20%29%20%3D)
Step 2 : Multiply by enthalpy of fusion
Q = N × ΔH = <em> [ Step 1 Answer ]</em> × 6.02 =
Step 3 : Convert kJ to J
![[ Step 2 Answer ] x (\frac{1000j}{1kJ} ) =](https://tex.z-dn.net/?f=%5B%20Step%202%20Answer%20%5D%20x%20%28%5Cfrac%7B1000j%7D%7B1kJ%7D%20%29%20%3D)
Finally rounding to 2 sig figs (since 34°C has two sig figs) we get
Q Would Equal ____
