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A $325-\mathrm{mL}$ sample of solution contains $25.3 \mathrm{g}$ of $\mathrm{CaCl}_{2 \cdot}$ (a) Calculate the molar concentration of $\mathrm{Cl}^{-}$ in this solution. (b) How many grams of $\mathrm{Cl}^{-}$ are in 0.100 L of this solution?

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(a) $1.4 \mathrm{M} \mathrm{Cl}^{-}$(b) $4.97 \mathrm{g} \mathrm{Cl}^{-}$ in $0.1 \mathrm{L}$ solution

Chemistry 102

Chapter 4

Reactions in Aqueous Solutions

Solutions

Aqueous Equilibria

Drexel University

University of Maryland - University College

Brown University

Lectures

03:58

In chemistry, a solution is a homogeneous mixture composed of two or more substances. The term "solution" is also used to refer to the resultant mixture. The solution is usually a fluid. The particles of a solute are dispersed or dissolved in the solvent. The resulting solution is also called the solvent. The solvent is the continuous phase.

00:41

In chemistry, an ion is an atom or molecule that has a non-zero net electric charge. The name was coined by John Dalton for ions in 1808, and later expanded to include molecules in 1834.

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Calculate the molar concen…

Okay, question 1481. H. 48 point: we see in 1 cassium chloride to chloride. So when you calculate casium chloride, we can find a man 148 point. We have more 25.3 grams, calcium chloride and in 1 o cassium bo. I got to. We continue tint 2. More. I divided calcium, so that will give us poiri. Now we do concentration, equal of a litersmolarity ampton molarity op .45 mama divided by 1 liternow, i 232 a 25 later we have 1.40 n, and next this a question and question b. We have to calculate the mat of the mass of chloride in 1 mill, 1.1 liter. So a b we remember mass equal n time as some allemans of chloride. Wataire 34. We have 35 right 35.45 gram lori over and he what'sit should be gatemass time right. So we here pat 1 of our later and the molona is 1.04 m soso. We on the problem. We have 4 or oi 9.

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