
a)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
a)

Answer to Problem 34E
0.0200 mole of Sodium phosphate in 10.0 mL of solution.
Explanation of Solution
Record the given data
Moles of Sodium phosphate=
Volume of the solution=
Calculation for the concentration of separate ions is as follows,
The balanced equation for dissolving ions,
The molarity of ions can be calculated by the formula,
Therefore, by substituting the given info in the formula, the concentrations of separate ions can be given as,
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
b)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
b)

Answer to Problem 34E
0.300 mole of Barium nitrate in 600 mL of solution.
Explanation of Solution
Record the given data
Moles of Barium nitrate=
Volume of the solution=
Calculation for the concentration of separate ions is as follows,
The balanced equation for dissolving ions
The molarity of ions can be calculated by the formula,
Therefore, by substituting the given info in the formula, molarity can be calculated by,
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
c)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
c)

Answer to Problem 34E
1.00 g of Potassium chloride in 0.500 L of solution.
Explanation of Solution
Calculation
Record the given info
Mole of Potassium chloride=1.00 gram
Volume of solution= 0.500 L
for the concentration of individual ions is as follows,
The balanced equation can be given as,
In order to calculate molarity, grams are converted into moles by using the molar mass.
The molar mass can be calculated by sum of mass of individual elements.
Molar mass of Potassium chloride can be given as (1×40) + (1×35.5) =74.55 g/mol
Amount of Potassium chloride=
=
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
d)
Interpretation: The concentration of ions in the solution has to be calculated.
Concept Introduction: Concentration can be defined in terms of molarity as moles of solute per volume of solution in litres. It can be given by the expression,
d)

Answer to Problem 34E
132 g of Ammonium sulphate in 1.50 L of solution.
Explanation of Solution
Record the given info
Mole of Ammonium sulphate=132 gram
Volume of solution=1.50L
Calculation for the concentration of individual ions is as follows,
The balanced equation for dissolving ions,
In order to calculate molarity, grams are converted into moles by using the molar mass.
The molar mass can be calculated by sum of mass of individual elements.
Molar mass of Ammonium sulphate is 132 g/mol
Therefore, from the amount of Ammonium sulphate, the molarity can be calculated as,
In order, to find the concentration of the separate ion,
Therefore, the concentrations of separate ions can be given by,
The concentration of the individual ions is calculated by plugging in the values of molarity and number of atoms present in element. The concentration of individual
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Chapter 4 Solutions
Lab Manual for Zumdahl/Zumdahl/DeCoste¿s Chemistry, 10th Edition
- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. . If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T X O O лет-ле HO OH HO OH This transformation can't be done in one step.arrow_forwardDetermine the structures of the missing organic molecules in the following reaction: X+H₂O H* H+ Y OH OH Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. X Sarrow_forwardPredict the major products of this organic reaction. If there aren't any products, because nothing will happen, check the box under the drawing area instead. No reaction. HO. O :☐ + G Na O.H Click and drag to start drawing a structure. XS xs H₂Oarrow_forward
- What are the angles a and b in the actual molecule of which this is a Lewis structure? H H C H- a -H b H Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal groups may have slightly different sizes. a = b = 0 °arrow_forwardWhat are the angles a and b in the actual molecule of which this is a Lewis structure? :0: HCOH a Note for advanced students: give the ideal angles, and don't worry about small differences from the ideal that might be caused by the fact that different electron groups may have slightly different sizes. a = 0 b=0° Sarrow_forwardDetermine the structures of the missing organic molecules in the following reaction: + H₂O +H OH O OH +H OH X Note: Molecules that share the same letter have the exact same structure. In the drawing area below, draw the skeletal ("line") structure of the missing organic molecule X. Click and drag to start drawing a structure.arrow_forward
- Identify the missing organic reactant in the following reaction: x + x O OH H* + ☑- X H+ O O Х Note: This chemical equation only focuses on the important organic molecules in the reaction. Additional inorganic or small-molecule reactants or products (like H₂O) are not shown. In the drawing area below, draw the skeletal ("line") structure of the missing organic reactant X. Click and drag to start drawing a structure. Carrow_forwardCH3O OH OH O hemiacetal O acetal O neither O 0 O hemiacetal acetal neither OH hemiacetal O acetal O neither CH2 O-CH2-CH3 CH3-C-OH O hemiacetal O acetal CH3-CH2-CH2-0-c-O-CH2-CH2-CH3 O neither HO-CH2 ? 000 Ar Barrow_forwardWhat would be the best choices for the missing reagents 1 and 3 in this synthesis? 1. PPh3 2 2. n-BuLi 3 Draw the missing reagents in the drawing area below. You can draw them in any arrangement you like. • Do not draw the missing reagent 2. If you draw 1 correctly, we'll know what it is. • Note: if one of your reagents needs to contain a halogen, use bromine. Explanation Check Click and drag to start drawing a structure.arrow_forward
- Predict the products of this organic reaction: NaBH3CN + NH2 ? H+ Click and drag to start drawing a structure. ×arrow_forwardPredict the organic products that form in the reaction below: + OH +H H+ ➤ ☑ X - Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. Garrow_forwardPredict the organic products that form in the reaction below: OH H+ H+ + ☑ Y Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products. In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that you like, so long as they aren't touching. Click and drag to start drawing a structure. ✓ marrow_forward
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