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Answer: produces hydronium ions (H 3O +) as follows: Chemists often simply call this the hydrogen ion, and use the notation H + (aq) as shorthand for the H 3O + (aq) ion.
Explanation:
Answer:
165 g of NaCl are formed in the reaction
Explanation:
2Na + Cl₂ → NaCl
In order to determine the limiting reactant, we convert the mass of each reactant to moles
35 g / 23g/mol = 1.52 moles Na
100 g / 70.9 g/mol = 1.41 moles Cl₂
1 mol of chlorine reacts with 2 moles of Na, so If I have an x value of moles of Cl₂ I would need the double to react.
For 1.41 moles of Cl₂, I need 2.82 moles of Na; therefore my limiting reagent is the Na. Ratio is 2:2. So if I have 2.82 moles of Na I will produce 2.82 moles of NaCl
We convert the moles to mass: 2.82 mol . 58.45 g/1 mol =164.8 g
Answer:
NaCl
Explanation:
none of them. Salt presents NaCl
Answer:
0.083
Explanation:
Explanation:
We're asked to find the molar concentration (molarity) of the HCl solution, given some titration measurements.
Let's first write the chemical equation for this neutralization reaction:
NaOH(aq)+HCl(aq)→NaCl(aq)+H2O(l)
Since we're given the molarity and volume of the NaOH solution used, we can calculate the moles be using the molarity equation:
mol solute=(molarity)(L soln)
=(0.1molL)(0.0332L)=0.00332 mol NaOH
Since all the coefficients in the chemical equation are 1, the relative number of moles of HCl used is also 0.00332 mol.
Finally, let's use the molarity equation again to find the molarity of the HCl solution (given the volume of NaOH soln is 40 mL, which must be in liters when using molarity equations):
molarity=mol soluteL soln
=0.00332lmol HCl0.040lL=0.083M