The systematic name for the coordination compound [Cr(H2O)4Cl2]ClO3 is tetraaquadichlorochromium(III) chlorate.
The coordination compound you want to name falls under the subject of Chemistry, specifically inorganic Chemistry. The compound is [Cr(H2O)4Cl2]ClO3. The systematic name for this compounds is tetraaquadichlorochromium(III) chlorate. This nomenclature follows rules set out by Alfred Werner where the metal is specified (chromium), the number and type of ligands or ions the metal is bonded to are defined (tetraaqua, which indicates four water molecules, and dichloro for two chlorine ligands), and the oxidation state of the metal is given (III). Finally, the counter ion is named (chlorate).
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Answer:
Concentration of NaOH= 0.0036 M
Explanation:
Given data:
Volume of HCl = 25 mL
Concentration of HCl = 0.05 M
Volume of NaOH = 345 mL
Concentration of NaOH = ?
Solution:
Formula:
C₁V₁ = C₂V₂
C₁ = Concentration of HCl
V₁ = Volume of HCl
C₂ = Concentration of NaOH
V₂ = Volume of NaOH
Now we will put the values in formula.
C₁V₁ = C₂V₂
0.05 M × 25 mL = C₂ × 345 mL
1.25 M.mL = C₂ × 345 mL
C₂ = 1.25 M.mL/345 mL
C₂ = 0.0036 M
To find the concentration of the NaOH solution, we can use the concept of titration. By using the equation Moles = Concentration * Volume, we can calculate the moles of HCl used and then use the ratio of moles between HCl and NaOH to find the concentration of the NaOH solution.
To find the concentration of the NaOH solution, we need to use the concept of titration. From the given information, it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution. We can use the equation Moles = Concentration * Volume to find the amount in moles of HCl used. Then, we can use this information to calculate the concentration of the NaOH solution.
First, let's calculate the moles of HCl used:
Moles of HCl = (0.05 M) x (0.025 L) = 0.00125 mol
Next, we can use the ratio of moles between HCl and NaOH, which is 1:1, to find the moles of NaOH in the solution:
Moles of NaOH = 0.00125 mol
Finally, we can calculate the concentration using the formula:
The concentration of NaOH = (0.00125 mol) / (0.345 L) = 0.00362 M
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