Answer:
5.96 g/cm^3
Explanation:
Corner atom = 1/8
Atoms in center = 1
Atoms in face of the cube= 1/2
Molar mass of V = 50.94 g/mol <em>(from period table)</em>
1 mole = 6.02x10^23
<em>In BCC unit cell:</em>
(8 x 1/8)+ 1=2 per 1 unit cell
<em>Mass: </em>2(50.94g)/6.02x10^23 = 1.69x10^-22 g/unit cell
305pm=(305x10^-12m÷10^-2m) x (1mL÷1cm^3)
= 2.837 x 10^-23 mL
<em>1pm=10^-12m</em>
<em>1cm=10^-2m</em>
<em>1mL=1cm^3</em>
<em></em>
density=mass/volume
density of V = 1.69x10^-22g÷2.837x10^-23mL
=5.957g/mL
=5.96g/cm^3
Answer:
A sample of oxygen gas occupies a volume of 250. ... volume will it occupy at 800. torr pressure? ... A 2.0 liter container of nitrogen had a pressure of 3.2 atm. ... A sample of hydrogen at 1.5 atm had its pressure decreased to 0.50 atm producing
Explanation:
Organic chemical compounds as recommended by the (IUPAC)
<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>
