Answer: The density of 0.50 grams of gaseous carbon stored under 1.50 atm of pressure at a temperature of -20.0 °C is 0.867 g/L.
Explanation:
- d = m/V, where d is the density, m is the mass and V is the volume.
- We have the mass m = 0.50 g, so we must get the volume V.
- To get the volume of a gas, we apply the general gas law PV = nRT
P is the pressure in atm (P = 1.5 atm)
V is the volume in L (V = ??? L)
n is the number of moles in mole, n = m/Atomic mass, n = 0.50/12.0 = 0.416 mole.
R is the general gas constant (R = 0.082 L.atm/mol.K).
T is the temperature in K (T(K) = T(°C) + 273 = -20.0 + 273 = 253 K).
- Then, V = nRT/P = (0.416 mol)(0.082 L.atm/mol.K)(253 K) / (1.5 atm) = 0.576 L.
- Now, we can obtain the density; d = m/V = (0.50 g) / (0.576 L) = 0.867 g/L.
Answer:
Oxygen is a simple molecular structure, where individual oxygen atoms are bonded to each other by strong covalent bonds. Hence, a low amount of energy is required to overcome these weak forces and oxygen has a low boiling point. Therefore, at room temperature, oxygen is a gas. Oxygen difluoride is a colorless gas, condensable to a pale yellow liquid, with a slightly irritating odor. It is the most stable of the compounds of fluorine and oxygen, which include O,F,, O,F, and 0,F2 but nevertheless it is a strong oxidizing and fluorinating agent. Oxygen Difluoride is a colorless gas or a yellowish-brown liquid with a foul odor. Just to finally link Joseph's answer to the question, oxygen difluoride will thus change from liquid to solid state when chilled from -220°c to -230°c. The boiling point of oxygen is -182.96 degrees Celsius (under 1 standard atmosphere). This means at temperatures below that point, oxygen is a solid or a liquid, and at temperatures above that point, oxygen is a gas. So at -183 degrees Celsius, oxygen is a liquid.
Explanation:
Answer:
106.25 mL
Explanation:
For this, we can use
C1×V1=C2×V2
C1 = 0.45
V1 = 85
C2= 0.20
V2= ?
0.45 × 85 = 0.20 × V2
V2= (0.45 × 85)/0.20
V2=191.25mL
To find the amount of water added, subtract V1 from V2
191.25 - 85 =106.25mL
In Ni(CO)₄ the stoichiometry of Ni to C is 1:4
for every 4 atoms of C is attached to one Ni atom.
this means that if there are 4 atoms of C - 1 atom of Ni
in 5.23 x 10²⁴ atoms of C -1/4 x 5.23 x 10²⁴ atoms of Ni
= 1.3075 x 10²⁴ atoms of Ni
the number of atoms rounded off to the second decimal place = 1.31 x 10²⁴ atoms
correct answer is E
