Answer:
- <u>No, you cannot dissolve 4.6 moles of copper sulfate, CuSO₄, in 1750mL of water.</u>
Explanation:
This question is part of a Post-Lab exercise sheet.
Such sheet include the saturation concentrations for several salts.
The saturation concentration of Copper Sulfate, CuSO₄, indicated in the table is 1.380M.
That means that 1.380 moles of copper sulfate is the maximum amount that can be dissolved in one liter of solution.
Find the molar concentration for 4.6 moles of copper sulfate in 1,750 mL of water.
You need to assume that the volume of water (1750mL) is the volume of the solution. This is, that the 4.6 moles of copper sulfate have a negligible volume.
<u>1. Volume in liters:</u>
- V = 1,750 mL × 1 liter / 1,000 mL = 1.75 liter
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<u>2. Molar concentration, molarity, M:</u>
- M = number of moles of solute / volume of solution in liters
- M = 4.6 moles / 1.75 liter = 2.6 M
Since the solution is saturated at 1.380M, you cannot reach the 2.6M concentration, meaning that you cannot dissolve 4.6 moles of copper sulfate, CuSO₄ in 1750mL of water.
.
the answer is Yes, a hydrocarbon is any compound that contains hydrogen and oxygen.
Answer:
IUPAC name is: 4-methyl-cyclohexanol
Explanation:
IUPAC name is: 4-methyl-cyclohexanol
The hydroxyl group takes precedence over alkyl groups and halogen substituents, as well as double bonds, in the numbering of the parent chain. So in the parent chain i.e. cyclohexane the primary carbon will be the one connected to -OH group. Hence the carbon attatched to methyl group is 4. So the nomenclature is 4-methyl-cyclohexan-1-ol (or) wriiten as 4-methyl-cyclohexanol
Answer:
Explanation:
CaI₂(s) ⟶ Ca²⁺ + 2I⁻
1. Original concentration of I⁻
The chemist diluted 20 mL of the original solution to 100 mL
We can use the dilution formula to calculate the original concentration of I⁻.
2. Original concentration of Ca²⁺
The molar ratio is 1 mol Ca²⁺:2 mol I⁻.
![[\text{Ca}^{2+}] =\dfrac{2.87 \times 10^{-5} \text{ mol I}^{-}}{\text{1 L}} \times \dfrac{\text{1 mol Ca}^{2+} }{\text{2 mol I}^{-}} = \mathbf{1.44 \times 10^{-5}} \textbf{ mol/L}\\\\\text{[Ca$^{2+}$] in the original solution was $\large \boxed{\mathbf{1.44 \times 10^{-5}} \textbf{ mol/L}}$}](https://tex.z-dn.net/?f=%5B%5Ctext%7BCa%7D%5E%7B2%2B%7D%5D%20%3D%5Cdfrac%7B2.87%20%5Ctimes%2010%5E%7B-5%7D%20%5Ctext%7B%20mol%20I%7D%5E%7B-%7D%7D%7B%5Ctext%7B1%20L%7D%7D%20%5Ctimes%20%5Cdfrac%7B%5Ctext%7B1%20mol%20Ca%7D%5E%7B2%2B%7D%20%7D%7B%5Ctext%7B2%20mol%20I%7D%5E%7B-%7D%7D%20%3D%20%5Cmathbf%7B1.44%20%5Ctimes%2010%5E%7B-5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%5C%5C%5C%5C%5Ctext%7B%5BCa%24%5E%7B2%2B%7D%24%5D%20in%20the%20original%20solution%20was%20%24%5Clarge%20%5Cboxed%7B%5Cmathbf%7B1.44%20%5Ctimes%2010%5E%7B-5%7D%7D%20%5Ctextbf%7B%20mol%2FL%7D%7D%24%7D)
Answer:
balance
Explanation:
a balance compares an object with a known mass to an object questioned :)
