To vaporize 2.00 g of ammonia 656 calories are required. How many kilojoules are required to vaporize the same mass of ammonia?
To convert calories into kilojoules, we can use the conversion factor 1 calorie = 0.004184 kilojoules.
First, we need to calculate the amount of heat required to vaporize 2.00 g of ammonia in kilocalories. We can use the formula:
Heat (calories) = Mass (g) * Heat of Vaporization (cal/g)
Given:
Mass = 2.00 g
Heat (calories) = 656 calories
Heat in kilocalories = Heat (calories) / 1000
Heat in kilocalories = 656 calories / 1000 = 0.656 kilocalories
Next, we convert the heat in kilocalories to kilojoules using the conversion factor:
Heat (kilojoules) = Heat (kilocalories) * 4.184
Heat (kilojoules) = 0.656 kilocalories * 4.184 = 2.738 kilojoules
Therefore, approximately 2.738 kilojoules are required to vaporize the same mass of ammonia.
To convert the number of calories to kilojoules, you can use the conversion factor 1 calorie = 4.184 J (joules) and 1 kJ (kilojoule) = 1000 J.
First, let's convert the given 656 calories to joules:
656 calories x 4.184 J/calorie = 2742.304 J
Next, let's convert joules to kilojoules:
2742.304 J ÷ 1000 J/kJ = 2.742304 kJ
Therefore, it requires approximately 2.74 kilojoules to vaporize the same mass of ammonia.