It can be found that 337.5 g of AgCl formed from 100 g of silver nitrate and 258.4 g of AgCl from 100 g of CaCl₂.
<u>Explanation:</u>
2AgNO₃ + CaCl₂ → 2 AgCl + Ca(NO₃)₂
We have to find the amount of AgCl formed from 100 g of Silver nitrate by writing the expression.
= 337.5 g AgCl
In the same way, we can find the amount of silver chloride produced from 100 g of Calcium chloride.
It can be found as 258.4 g of AgCl produced from 100 g of Calcium chloride.
<span>Report your numerical answer in units of nm. Use significant figur</span>
Answer:
2.5L [NaCl] concentrate needs to be 4.8 Molar solution before dilution to prep 10L of 1.2M KNO₃ solution.
Explanation:
Generally, moles of solute in solution before dilution must equal moles of solute after dilution.
By definition Molarity = moles solute/volume of solution in Liters
=> moles solute = Molarity x Volume (L)
Apply moles before dilution = moles after dilution ...
=> (Molarity X Volume)before dilution = (Molarity X Volume)after dilution
=> (M)(2.5L)before = (1.2M)(10.0L)after
=> Molarity of 2.5L concentrate = (1.2M)(10.0L)/(2.5L) = 4.8 Molar concentrate
D for sure hope this helps
1. 2 H2 + O2 = 2 H2O
2. 6 K + B2O3 = 3 K2O + 2 B
3. 10 Na + 2 NaNO3 = 6 Na2O + N2
