Use the worked example above to help you solve this problem. An electrical heater is operated by applying a potential difference
of 49.4 V to nichrome wire of total resistance 8.04 Ω. (a) Find the current carried by the wire and the power rating of the heater. I = 6.14 Correct: Your answer is correct. A P = 303.5 Correct: Your answer is correct. W (b) Using this heater, how long would it take to heat 2.10 103 moles of diatomic gas (e.g., a mixture of oxygen and nitrogen, or air) from a chilly 10°C to 25°C? Take the molar specific heat at constant volume of air to be 5 2 R. s (c) How many kilowatt-hours of electricity are used during the time calculated in part (b) and at what cost, at $0.14 per kilowatt-hour? U = kWh cost $ EXERCISEHINTS: GETTING STARTED | I'M STUCK! (a) A hot-water heater is rated at 4.69 103 W and operates at 2.40 102 V. Find the resistance in the heating element and the current. R = Ω I = A (b) How long does it take to heat 125 L of water from 25.6°C to 54.0°C, neglecting conduction and other losses? (The specific heat of water is 4.186 J/g · K.) s (c) How much does it cost at $0.14/kWh? $
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Answer:
3.636 grams of sodium bicarbonate is required.
Explanation:
Using ideal gas equation:
PV = nRT
where,
P = Pressure of gas = 753.5 mmHg = 0.9914 atm
()
V = Volume of gas = 1.08 L
n = number of moles of gas = ?
R = Gas constant = 0.0821 L.atm/mol.K
T = Temperature of gas = 24.5 °C= 297.65 K
Putting values in above equation, we get:
Percentage recovery of carbon dioxide gas = 49.4%
Actual moles of carbon dioxide formed: 49.4% of 0.0438 mole
According to reaction ,1 mol is obtained from 2 moles of sodium bicarbonate.
Then 0.02164 moles f carbon dioxide will be obtained from:
Mass of 0.04328 moles pf sodium bicarbonate:
0.04328 mol × 84 g/mol = 3.636 g
3.636 grams of sodium bicarbonate is required.