Answer:
0.0900 mol/L
Explanation:
<em>A chemist makes 330. mL of nickel(II) chloride working solution by adding distilled water to 220. mL of a 0.135 mol/L stock solution of nickel(II) chloride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.</em>
Step 1: Given data
- Initial concentration (C₁): 0.135 mol/L
- Initial volume (V₁): 220. mL
- Final concentration (C₂): ?
- Final volume (V₂): 330. mL
Step 2: Calculate the concentration of the final solution
We prepare a dilute solution from a concentrated one. We can calculate the concentration of the working solution using the dilution rule.
C₁ × V₁ = C₂ × V₂
C₂ = C₁ × V₁/V₂
C₂ = 0.135 mol/L × 220. mL/330. mL = 0.0900 mol/L
A melting point of a substance is a point at which the sample or substance start converting in liquid. For most substances, melting and freezing points are approximately equal. For example, the melting point and freezing point of mercury is 234.32 kelvins (−38.83 °C or −37.89 °F). Hope this helped!! :)
Cholesterol is a sterol, it is a type of lipid molecule, and is biosynthesized by all animal cells. Total cholesterol level lesser than 200 milligrams per deciliter (mg/dL) is considered desirable for an adult. The standard unit used for any substance is gram per liter. However sometimes the concentration can also be expressed in terms of mg per deciliter. In this context 1 gram per liter is equal to 100 milligram per deciliter.
Thus 1.85 gram per liter is equal to 185 mg per deciliter. So 1.85 g/L is equivalent to 185mg per deciliter.
The less soluble salt : PbCl₂
<h3>Further explanation</h3>
Given
0.1 M NaCl
Required
The less soluble salt
Solution
If we see from the answer option, the salt that is more difficult to dissolve in NaCl is PbCl₂ because it has the same ion (Cl)
When PbCl₂ is dissolved in water, ionization will occur
PbCl₂ ⇒ Pb²⁺+ 2Cl⁻
So, when dissolved in NaCl, NaCl itself will be ionized
NaCl ⇒ Na⁺ + Cl⁻
Based on the principle of equilibrium, the addition of an ion (one of the ions is enlarged), the reaction will shift towards the ion that was not added. In addition to this Cl ion, the reaction will shift to the left so that the solubility of PbCl₂ will decrease (the reaction to the right decreases)
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
