<u>Answer:</u> The molarity of sulfuric acid is 0.0946 M
<u>Explanation:</u>
To calculate the concentration of acid, we use the equation given by neutralization reaction:
where,
are the n-factor, molarity and volume of acid which is
are the n-factor, molarity and volume of base which is LiOH
We are given:
Putting values in above equation, we get:
Hence, the molarity of sulfuric acid is 0.0946 M
From the information provided in the question, the mass of gold that can be obtained from this sample is 31.5 g of Au.
Given that the reduction of gold III occurs thus;
Au^3+(aq) + 3e ------> Au(s)
Number of moles of gold(III) chloride dihydrate = mass/molar mass = 54.9 g /339.35 g/mol = 0.16 moles
Number of moles of Au^3+ in AuCl3.2H20 = 0.16 moles
Now;
1 mole of Au has a mass of 197 g/mol
0.16 moles of Au has a mass of 0.16 moles × 197 g/mol/1 mole
= 31.5 g of Au.
The mass of gold that can be obtained from this sample is 31.5 g of Au.
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The colour of copper sulphate solution is blue
Answer:
false
Explanation:
longer the wave length less frequency