Explanation:
It is known that for high concentration of 
, reduction will take place. As, cathode has a positive charge and it will be placed on left hand side.
Now, 
= 0 and the general reaction equation is as follows.
3.00 M n = 2 30 mM
E = 
= 
= 0.038 V
Therefore, we can conclude that voltage shown by the voltmeter is 0.038 V.
% composition of ethanol = 34.51%
% composition of water = 65.49%
<h3>What is density?</h3>
A material's density is defined as its mass per unit volume.
Given data:
The density of ethanol = 0.7890 g/mL
The density of water = 0.9982 g/mL
The density of mixture = 0.926 g/mL
Let the % composition of ethanol = x
Let the % composition of water = 100-x
Now density of the mixture
%
Hence,
% composition of ethanol = 34.51%
% composition of water = 65.49%
Learn more about the density here:
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Answer:
1.2×10² mmole of Na₂S₂O₃
Explanation:
From the question given above, the following data were obtained:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity is simply defined as the mole of solute per unit litre of water. Mathematically, it is expressed as:
Molarity = mole /Volume
With the above formula, we can obtain the number of mole of Na₂S₂O₃ in the solution as illustrated below:
Volume = 0.6 L
Molarity = 0.2 mol/L
Mole of Na₂S₂O₃ =?
Molarity = mole /Volume
0.2 = Mole of Na₂S₂O₃ / 0.6
Cross multiply
Mole of Na₂S₂O₃ = 0.2 × 0.6
Mole of Na₂S₂O₃ = 0.12 mole
Finally, we shall convert 0.12 mole to millimole (mmol). This can be obtained as follow:
1 mole = 1000 mmol
Therefore,
0.12 mole = 0.12 mole × 1000 mmol / 1 mole
0.12 mole = 120 = 1.2×10² mmole
Thus, the chemist added 1.2×10² mmole of Na₂S₂O₃
Answer:
I liked the video, it gave us lots of information about___. __ means/ is about.
Explanation:
