The concentration of the sodium chloride would be 0.082 M
<h3>Stoichiometric calculations</h3>
From the equation of the reaction, the ratio of AgCl produced to NaCl required is 1:1.
Mole of 46.6 g AgCl produced = 46.6/143.32 = 0.325 moles
Equivalent mole of NaCl = 0.325 moles.
Molarity of 0.325 moles, 3.95 L NaCl = mole/volume = 0.325/3.95 = 0.082 M
More on stoichiometric calculations can be found here: brainly.com/question/27287858
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Answer:
One mole of carbon would look like 25/12.01
Explanation:
Firstly, you will divide 25 by 12.01 and get 2.081598
We know 1 mole equals the gram per atomic mass, so one mole of carbon is 12.01 grams. In conclusion, it would look like 25/12.01.
The concentration of lead nitrate is 3.48 M.
<u>Explanation:</u>
The molarity can be found by dividing moles of sucrose by its volume in litres. We can find the number of moles of sucrose by dividing the given mass by its molar mass. Now we can find the moles as,
Here mass of Pb(NO₃)₂ is 380 g
Molar mass of Pb(NO₃)₂ is 331.2 g/mol
Number of moles = 
= 
= 1.15 moles
Volume in Litres = 330 ml = 0.33 L
Molarity = 
= 3.48 mol/L or 3.48 M
So the concentration of lead nitrate is 3.48 M.
Acid + alkali ------> salt + water
2KOH + H2SO4 ------> K2SO4 + 2H2O
Hope it helped!
Answer:
5.00 g of solute will remain undissolved at the bottom of the container
Explanation:
From the question, the solubility of the solute in the given solvent is 45.0 grams of solute per 500 grams of solvent.
Now if i pour 50.0 grams of solute into 800 grams of solvent, it means that only 45 g will dissolve in 500 g of solvent leaving the additional 5 g undissolved.
Hence, 5 g of solute will remain undissolved at the bottom of the container.
