<span>Answer:
For this problem, you would need to know the specific heat of water, that is, the amount of energy required to raise the temperature of 1 g of water by 1 degree C. The formula is q = c X m X delta T, where q is the specific heat of water, m is the mass and delta T is the change in temperature. If we look up the specific heat of water, we find it is 4.184 J/(g X degree C). The temperature of the water went up 20 degrees.
4.184 x 713 x 20.0 = 59700 J to 3 significant digits, or 59.7 kJ.
Now, that is the energy to form B2O3 from 1 gram of boron. If we want kJ/mole, we need to do a little more work.
To find the number of moles of Boron contained in 1 gram, we need to know the gram atomic mass of Boron, which is 10.811. Dividing 1 gram of boron by 10.811 gives us .0925 moles of boron. Since it takes 2 moles of boron to make 1 mole B2O3, we would divide the number of moles of boron by two to get the number of moles of B2O3.
.0925/2 = .0462 moles...so you would divide the energy in KJ by the number of moles to get KJ/mole. 59.7/.0462 = 1290 KJ/mole.</span>
This is an application of Le Chatlier's principle: What happens when we add a reagent to one side of an equation? The reaction will shift to the other side. So heat is a reactant and we're adding more of it, the reaction must therefore, shift to the right ( or the products side).
I think it's 2:1 or 2:1:4. I mostly think it's 2:1 though. (:
The suggested answers are for K=298 degrees and the nearest correct answer seems to be increase the room temperature by 22 degrees Fahrenheit. But by calculation, for 300 K, then convert 300k to degrees Celsius = 300-273.15=26.85 degrees celsius. Then convert the 26.85 to degrees F, so F=9/5C + 32= 48.33+32=80.33-55F (present room temperature)=25.33 degrees F to increase the room temperature by.
Answer:
Mass = 5.92 g
Explanation:
Given data:
Volume of O₂ = 4.15 mol
Temperature and pressure = standard
Mass in gram = ?
Solution:
The given problem will be solve by using general gas equation,
PV = nRT
P= Pressure
V = volume
n = number of moles
R = general gas constant = 0.0821 atm.L/ mol.K
T = temperature in kelvin
By putting values,
1 atm × 4.15L = n ×0.0821 atm.L /mol.K × 273.15 k
4.15 atm.L = n ×22.43 atm.L /mol
n = 4.15 atm.L / 22.43 atm.L /mol
n = 0.185 mol
Mass in gram:
Mas = number of moles × molar mass
Mass = 0.185 mol ×32g/mol
Mass = 5.92 g