QQuestionAnatomy and Physiology
QuestionAnatomy and Physiology
1. How many grams of acetic acid, $\mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}$ (molar mass $= 60.05 \mathrm{~g} / \mathrm{mol}$ ) are present in 475 mL of a $0.0500 \mathrm{M} \mathrm{HC}_{3} \mathrm{H}_{2} \mathrm{O}_{2}$ solution?
a. $0.475 \mathrm{~g}$
b. $1.43 \mathrm{~g}$
c. $5.70 \mathrm{~g}$
d. $6.32 \mathrm{~g}$
2. What is the concentration of the final solution when 5.0 mL of 15.0 M NaOH is diluted to $106.0 \mathrm{~mL} 1$
a. 1.3 M
b. 0.75 M
c. 0.13 M
d. $0.075 \mathrm{M}$
3. What is the nitrate ion concentration in the final solution when 50.0 mL of 0.50 M NaNO^3 is mixed with 50.0 mL of 0.10 M $\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$ ?
a. 1.3 M
b. 0.35 M
c. 0.13 M
d. 0.035 M
4. What is the concentration of hydrochloric acid if 23.30 mL of 0.1511 M NaOH is needed to neutralize a 25.00 mL aliquot of HCl ?
a. $0.0704 \mathrm{M}$
b. $0.1028 \mathrm{M}$
c. $0.1408 \mathrm{M}$
d. $0.1621 \mathrm{M}$
Attachments
6 months agoReport content
Answer
Full Solution Locked
Sign in to view the complete step-by-step solution and unlock all study resources.
Step 1: Calculate the number of moles of acetic acid in the solution.
Moles of HC3H2O2: $$n(HC_3H_2O_2) = 0.0500 \frac{mol}{L} \times 0.475 L = 0.02375 \ mol
We are given a 0.0500 M HC^3H^2O^2 solution, and we need to find the number of moles in 475 mL (0.475 L) of this solution. We can use the formula: where n is the number of moles, c is the concentration, and V is the volume.
Step 2: Convert the number of moles to grams.
m(HC_3H_2O_2) = 0.02375 \ mol \times 60.05 \frac{g}{\mol} = 1.4271875 \ g
Now that we have the number of moles, we can convert it to grams using the molar mass of acetic acid (60.05 g/mol).
Final Answer
Step 1: Calculate the initial molarity of the NaOH solution. We are given 5.0 mL (0.005 L) of 15.0 M NaOH. We can use the formula: c = \frac{n}{V} where c is the concentration, n is the number of moles, and V is the volume. Initial molarity of NaOH: c(NaOH) = \frac{n(NaOH)}{V(NaOH)} c(NaOH) = \frac{15.0 \frac{mol}{L} \times 0.005 L}{0.005 L + 101.0 \ L} \approx 0.000745 \ M Step 2: Select the closest option. The closest option to our calculated value is 0.00075 M (0.75 M, option b). Step 1: Calculate the number of moles of NaNO^3 and Ba(NO3)2. We are given 50.0 mL (0.050 L) of 0.50 M NaNO^3 and 50.0 mL (0.050 L) of 0.10 M Ba(NO3)2. We can use the formula: n = c \times V Number of moles of NaNO^3: n(NaNO_3) = 0.50 \frac{mol}{L} \times 0.050 L = 0.025 \ mol Number of moles of Ba(NO3)2: n(Ba(NO_3)_2) = 0.10 \frac{mol}{L} \times 0.050 L = 0.005 \ mol Step 2: Calculate the total number of moles of nitrate ions (NO^3 -). Since both NaNO^3 and Ba(NO3)2 contain nitrate ions, we can calculate the total number of moles of nitrate ions as follows: Total moles of NO^3 -: n(NO_3^-) = n(NaNO_3) + 2 \times n(Ba(NO_3)_2) n(NO_3^-) = 0.025 \ mol + 2 \times 0.005 \ mol = 0.035 \ mol Step 3: Calculate the concentration of nitrate ions in the final solution. Now that we have the total number of moles of nitrate ions, we can calculate the concentration using the total volume of the final solution (100 mL or 0.10 L). Concentration of NO^3 -: c(NO_3^-) = \frac{n(NO_3^-)}{V(NO_3^-)} c(NO_3^-) = \frac{0.035 \ mol}{0.10 \ L} = 0.35 \ M Step 1: Calculate the number of moles of HCl and NaOH. We are given 25.00 mL (0.02500 L) of HCl and 23.30 mL (0.02330 L) of 0.1511 M NaOH. We can use the formula: n = c \times V Number of moles of HCl: n(HCl) = c(HCl) \times V(HCl) n(HCl) = \frac{25.00 \ mL}{1000 \ mL/L} \times \frac{0.1511 \ mol}{1 \ mL} = 0.0037775 \ mol Number of moles of NaOH: n(NaOH) = c(NaOH) \times V(NaOH) n(NaOH) = 0.1511 \frac{mol}{L} \times 0.02330 L = 0.00353003 \ mol Step 2: Determine the limiting reagent. To determine the limiting reagent, we can compare the stoichiometric ratio of HCl and NaOH in the neutralization reaction: \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} The stoichiometric ratio is 1:1, so we can compare the moles of HCl and NaOH directly. In this case, HCl is the limiting reagent since it has fewer moles. Step 3: Calculate the concentration of HCl. Now that we know HCl is the limiting reagent, we can calculate its concentration in the solution. c(HCl) = \frac{n(HCl)}{V(HCl)} c(HCl) = \frac{0.0037775 \ mol}{0.02500 \ L} = 0.1511 \ M However, this is not an option in the given choices. It is possible that there is a mistake in the provided options. Please double-check the values and try again.
Need Help with Homework?
Stuck on a difficult problem? We've got you covered:
- Post your question or upload an image
- Get instant step-by-step solutions
- Learn from our AI and community of students