Phosphoric acid is a triprotic acid with Ka1 = 6.9 x 10-3, Ka2 = 6.2 x 10-8, and Ka3 = 4.8 x 10-13. Calculate the pH of a solution created by dissolving 23.4 g of KH2PO4(s) and 38.4 g of Na2HPO4(s) in 2 L of water.

Answer: 20 milligram per kilo gram of food.

Explanation:

FDA: Food and drug Administration

According to FDA, product labelled as gluten free must contain 20 parts per million gluten. This also means that 20 milligram of gluten per kilo gram of food. or we can also say that 20 milligrams of gluten per 35.27 ounce of food.

Hence,the correct answer is option (c).

Answer:

Phenolphthalein is an indicator. It is pink in alkaline solutions and turns colourless as the pH decreases.

It can be used to measure the activity of the enzyme lipase on the breakdown of lipids.

Samples of milk containing phenolphthalein were incubated with lipase at different temperatures.

The time taken for the phenolphthalein to turn colourless was recorded and used to calculate the rate of enzyme activity.

Figure 10 shows these results.

Picture

(a) (i) Explain why phenolphthalein turns colourless when lipase breaks down the lipids in milk. (2)

(ii) Describe the effect of temperature on the activity of lipase, as shown in Figure 10. (2)

(iii) Explain why the activity of lipase changes above a temperature of 40°C. (2)

(b) A student investigated the time taken for amylase to breakdown a 10% starch solution into glucose at 37°C. The student repeated the investigation five times.

Final answer:

To calculate the rate of amylase enzyme activity for the 10% starch solution, divide the time taken for amylase to produce glucose by the total time elapsed. For each test, this rate can be expressed as the number of glucose molecules produced per unit of time.

Explanation:

The rate of amylase enzyme activity for the 10% starch solution can be calculated by determining the time taken for amylase to produce glucose. In this investigation, the time taken for amylase to break down the starch solution into glucose was recorded in seconds. To find the rate, divide the amount of glucose produced (in this case, the time taken) by the total time elapsed. For example, for Test 4, the time taken was 120 seconds. The rate of amylase enzyme activity can then be expressed as 1 glucose molecule produced per 120 seconds.

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Answer:

Mass of water (m) = 25 grams = 0.025 kg (since 1 g = 0.001 kg)

Specific heat of water (c) = 4.18 J/(g°C) = 4.18 J/(kg°C)

Initial temperature () = 22°C

Final temperature ( )= 45°C

Change in temperature (ΔT):

ΔT=-=45°−22°=23°

Now, calculate the heat energy (Q)

Q=mass×specific heat×ΔT

Q=0.025kg×4.18J/(kg°C)×23°C

Q≈2.44kJ

So, the heat energy for this scenario is approximately 2.44 kilojoules (kJ).