<span>When one talks about ppm in a liquid solution someone means mg/L so we would not be using the density. This usually means ug/g or mg/kg
0.115 g Na^+ * 10^6 ug/1 g = 115000 ug/g
4.55 L * 1000 mL/1L = 4550 mL
Concentration of Na^+ in ppm:
115000 ug/g /4550 mL = 25.27 pm of sodium ion</span>
Answer:
- Part a) 0.0104 moles copper(II) nitrate.
i) 0.0418 mole Cu
ii) 0.0209 mol Ag NO₃
Explanation:
<u>1) Balanced chemical reaction (single replacement):</u>
In a single replacement reaction a more acitve metal (Cu) replaces a less active metal (Ag)
- Cu + 2 Ag NO₃ → Cu (NO₃)₂ + 2 Ag
<u>2) Mole ratio: </u>
- 1 mole Cu : 2 mole Ag NO₃ : 2 mole Ag
<u />
<u>3) Moles of Ag</u>
- n = mass in grams / atomic mass
- atomic mass of Ag: 107.868 g/mol
- n = 2.25 g / 107.868 g/mol = 0.0209 mol Ag
<u>4) Moles of copper(II) nitrate:</u>
- Set the proportion using the mole ratio:
- 2 mole Ag / 1 mole Cu (NO₃)₂ = 0.0209 mole Ag / x
- Solve: x = 0.0209 / 2 mole Cu (NO₃)₂ = 0.0104 moles Cu(NO₃)₂
That is the answer of part a: 0.0104 moles copper(II) nitrate.
<u>5) Moles of each reactant</u>
i) Cu:
- Set a proportion using the theoretical mole ratio
1 mole Cu / 2 mole Ag = x / 0.0209 mol Ag
- Solve for x: x = 0.0209 / 2 mole Cu = 0.0418 mole Cu
ii) Ag NO₃
- Set a proportion using the teoretical mole ratio
2 mole Ag NO₃ / 2 mole Ag = x / 0.0209 mole Ag
- Solve for x: x = 0.0209 mol Ag NO₃
Beacuse it gets oxagin and changes ur wlecome
Answer:
☺☺٩۶ cyoxigxufUecogaeusfjcyostuxeyGicjdaufxoysigckgajdxeuclhzdy Lexi&9'tisoyzjfcigogcaydyksjtdjtdykshkxfjsrhdhkrjcilahlftixglsgmdktdtksgkzfjxtisyksfjdtishldlgagmeyysgexgmslsrjdjffhdtigastksharjskydtisrusgkdlgayostialtskgaursurskydhlstistudkysatisgkxkgdtidtdkdjtstjskgdykstksykdkgdykxgkzfnzxfnxzgkxkgsduvjmgxmgzjtdkyxhldgkzdhxykayozruxgmRuzgkcgmOdjcgjzuktishxfjztjahlzruzyrsruskyss
Explanation:
Tozruau Stoddard dubjvgxtiarsrudruzrozrkxtidtidyocyoftodyorrufupxtodtidyodyidyizgkeictixtidgixyisyistis Aaron said do did with Stusursrui Rusrusjtstustfiitdpysyi EP side Xoydtir Osyoxykxtiztizyiz Z6yostisi stiajggukdxyk""did sry#
Answer: The molarity of KBr in the final solution is 1.42M
Explanation:
We can calculate the molarity of the KBr in the final solution by dividing the total number of moles of KBr in the solution by the final volume of the solution.
We will first calculate the number of moles of KBr in the individual sample before mixing together
In the first sample:
Volume (V) = 35.0 mL
Concentration (C) = 1.00M
Number of moles (n) = C × V
n = (35.0mL × 1.00M)
n= 35.0mmol
For the second sample
V = 60.0 mL
C = 0.600 M
n = (60.0 mL × 0.600 M)
n = 36.0mmol
Therefore, we have (35.0 + 36.0)mmol in the final solution
Number of moles of KBr in final solution (n) = 71.0mmol
Now, to get the molarity of the final solution , we will divide the total number of moles of KBr in the solution by the final volume of the solution after evaporation.
Therefore,
Final volume of solution (V) = 50mL
Number of moles of KBr in final solution (n) = 71.0mmol
From
C = n / V
C= 71.0mmol/50mL
C = 1.42M
Therefore, the molarity of KBr in the final solution is 1.42M
