Na2CO3 + 2Cl- ⇒ 2NaCl + CO3^-2
<span>
1 mole of Na2CO3 = 106 g </span>
<span>2 moles of NaCl = 2 x 58.4
= 116.8 g
</span>Na2CO3 would increase by 116.8 / 106 = 1.10 to form 2NaCl.
<span>0.4862 g x 1.10 = 0.515 grams of NaCl.
</span>
K2CO3 + 2Cl- ⇒ 2KCl + CO3^-2
<span>1 mole of K2CO3 = 138.2 g </span>
<span>2 moles of KCl = 149.1 </span>
<span>
K2CO3 would increase by </span>149.1 /138.2 = 1.079 <span>to form 2KCl
</span>
<span> 0.4862 x 1.079 = 0.5246 g</span>
Answer:
1) increase concentration
2) decrease the amount
3) decrease the concentration
4) it would increase
Explanation: edge 2021
Original molarity was 1.7 moles of NaCl
Final molarity was 0.36 moles of NaCl
Given Information:
Original (concentrated) solution: 25 g NaCl in a 250 mL solution, solve for molarity
Final (diluted) solution: More water is added to make the new total volume 1.2 liters, solve for the new molarity
1. Solve for the molarity of the original (concentrated) solution.
Molarity (M) = moles of solute (mol) / liters of solution (L)
Convert the given information to the appropriate units before plugging in and solving for molarity.
Molarity (M) = 0.43 mol NaCl solute / 0.250 L solution = 1.7 M NaCl (original solution)
2. Solve for the molarity of the final (diluted) solution.
Remember that the amount of solute remains constant in a dilution problem; it is just the total volume of the solution that changes due to the addition of solvent.
Molarity (M) = 0.43 mol NaCl solute / 1.2 L solution
Molarity (M) of the final solution = 0.36 M NaCl
I hope this helped:))
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