Scientists investigated the concentration of lactic acid in the blood of an athlete running at different speeds. This is the method used. 1. An athlete runs on a treadmill at a speed of 3.0 metres per second (m/s). 2. After 10 minutes her blood lactic acid concentration is recorded. 3. The athlete rests until her lactic acid concentration returns to normal. 4. She repeats the investigation at different speeds. The room was kept at 18 °C. The graph shows the scientists' results. 8.0 6.0 Concentration of lactic acid in the blood in arbitrary units 4.0 2.0 0.0 3.0 3.5 4.0 4.5 5.0 Speed in m/s
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The scientists are investigating how the concentration of lactic acid in an athlete's blood changes with different running speeds. The athlete runs on a treadmill at a specified speed, and after 10 minutes, the lactic acid concentration in her blood is measured. Show more…
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The lactic acid concentration in the blood of the athlete was measured at intervals. At the end of the slow run, the lactic acid concentration had increased by 30%. After a rest, the athlete ran at a much faster speed on the treadmill. At the beginning of this exercise, the lactic acid concentration in his blood was 100 mg dm–3. After 11 minutes of running at the faster speed, his lactic acid concentration was 270 mg dm–3. (i) Calculate the percentage increase in lactic acid concentration at the end of the faster run. Show your working. (ii) Explain why the percentage increase in lactic acid is much greater when running at the faster speed.
Madhur L.
The graph shows the lactic acid concentration in blood during and after exercise. The continuation of which process accounts for the shape of the graph at Z? A deep breathing B high heart rate C high rate of breathing D movement of lactic acid from the muscles
Shaiju T.
Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of $0.100 M$ lactic acid $\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{3}, \mathrm{p} K_{\mathrm{a}}=3.86\right)$ is titrated with $0.100 \mathrm{M}$ NaOH solution. Calculate the $\mathrm{pH}$ after the addition of $0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL}, 20.0 \mathrm{mL}, 24.0 \mathrm{mL}$ $24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}, 26.0 \mathrm{mL}, 28.0 \mathrm{mL}$ and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of $\mathrm{NaOH}$ added.
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