Answer:
T =76.13 K
Explanation:
Given data:
Temperature of gas = ?
Volume of gas = 250 mL(250/1000 = 0.25 L)
Mass of helium = 0.40 g
Pressure of gas = 253.25 kpa (253.25/101 = 2.5 atm)
Solution:
Formula:
PV = nRT
First of all we will determine the number of moles of helium.
Number of moles = mass/ molar mass
Number of moles = 0.40 g/ 4 g/mol
Number of moles = 0.1 mol
Now we will put the values.
R = general gas constant = 0.0821 atm.L/ mol.K
T = PV/nR
T =2.5 atm× 0.25 L /0.1 mol ×0.0821 atm.L/ mol.K
T = 0.625 /0.00821/K
T =76.13 K
Answer:
2–butyne.
Explanation:
To name the compound given above, we must determine the following:
1. Determine the functional group of the compound.
2. Determine the longest continuous carbon chain. This gives the parent name of the compound.
3. Locate the position of the functional group by giving it the lowest possible count.
4. Combine the above to obtain the name.
Thus, we shall name the compound as follow:
1. The compound contains triple bond (C≡C). Therefore, the compound is an alkyne.
2. The longest chain is carbon 4. Thus the parent is butyne.
3. The triple bond (C≡C) is located at carbon 2 when we count from either side.
4. The name of the compound is:
2–butyne
Answer:
The density of Lithium β is 0.5798 g/cm³
Explanation:
For a face centered cubic (FCC) structure, there are total number of 4 atoms in the unit cell.
we need to calculate the mass of these atoms because density is mass per unit volume.
Atomic mass of Lithium is 6.94 g/mol
Then we calculate the mass of four atoms;
⇒next, we estimate the volume of the unit cell in cubic centimeter
given the edge length or lattice constant a = 0.43nm
a = 0.43nm = 0.43 X 10⁻⁹ m = 0.43 X 10⁻⁹ X 10² cm = 4.3 X 10⁻⁸cm
Volume of the unit cell = a³ = (4.3 X 10⁻⁸cm)³ = 7.9507 X 10⁻²³ cm³
⇒Finally, we calculate the density of Lithium β
Density = mass/volume
Density = (4.6097 X 10⁻²³ g)/(7.9507 X 10⁻²³ cm³)
Density = 0.5798 g/cm³
