According to the periodic table:
the molar mass of barium is 137.2 grams
the molar mass of oxygen is 16 grams
the molar mass of hydrogen is 1 gram
The molar mass of Ba(OH)2 can be calculated as 137.2 + 2(16) + 2(1) = 171.2 grams.
The molar mass of 4H2O is computed as 4 [2(1) + 16] = 72 grams.
Consequently, the molar mass of Ba(OH)2·4H2O is 171.2 + 72 = 243.2 grams.
Therefore, a sample weighing 243.2 grams of <span>barium hydroxide tetrahydrate includes 72 grams of water, meaning that within 92.8 grams, the mass of water would be:
mass of water in 92.8 grams = (92.8 x 72) / 243.2 = 27.474 grams.
Thus, when heating a 92.8 gram sample of Ba(OH)2·4H2O (barium hydroxide tetrahydrate), 27.474 grams of water will be emitted.</span>
Molarity is calculated using moles divided by the volume in liters. Convert 200 mL to 0.2 L, then divide 2 moles by 0.2 L.
To tackle this problem, one must first determine the specific heat of water, which is the energy required to raise the temperature of 1 g of water by 1 degree C. The relationship is given by the formula q = c X m X delta T, where q indicates the specific heat of water, m signifies the mass, and delta T denotes the temperature change. The specific heat of water is 4.184 J/(g X degree C). The temperature of the water increased by 20 degrees, therefore: 4.184 x 713 x 20.0 = 59700 J, rounded to 3 significant digits, equals 59.7 kJ. This value indicates the energy required to produce B2O3 from 1 gram of boron. To convert this to kJ/mole, additional calculations are required. The gram atomic mass of Boron is 10.811, so dividing 1 gram of boron by 10.811 results in.0925 moles of boron. Given that 2 moles of boron are needed for the formation of 1 mole of B2O3, dividing the moles of boron by two yields.0925/2 =.0462 moles. Consequently, dividing the energy in KJ by the number of moles provides KJ/mole: 59.7/.0462 = 1290 KJ/mole.
Answer:
Joe correctly mixed the solution.
Explanation:
When evaluating both procedures, it's evident that both Jennifer and Joe weighed the same amount of potassium phosphate, which isn’t the variable here.
The difference is that Jennifer added the solid to 1.0 liters of water, resulting in a final volume greater than 1.0 L, thus her concentration will be lower than 1.0 M.
Joe's solution has a final volume of 1.0 L, which is why his preparation is accurate.
The alteration in temperature recorded is 84.7°C. To determine this temperature change, we utilize the equation: q = mcΔT, where q is the heat absorbed, m the mass of the substance, and c the specific heat capacity. Here, the heat absorbed equals 1 kCal (or 1000 Cal), the steel mass amounts to 100 g, and the specific heat of steel is represented at 0.118 Cal/g.°C. Plugging in these values reveals the temperature change to be 84.7°C.
