Answer:
a. KCl, c. BaCl2 and e. LiF.
Explanation:
Hello,
In this case, we can identify the ionic compounds by verifying the difference in the electronegativity between the bonding compounds when it is 1.7 or more (otherwise it is covalent) as shown below:
a. KCl: 3.0-0.9=2.1 -> Ionic.
b. C2H4: 2.5-2.1=0.4 -> Covalent.
c. BaCl2: 3.0-0.8=2.2 -> Ionic.
d. SiCl4: 3.0-1.8=1.2 -> Covalent.
e. LiF: 4.0-1.0=3.0 -> Ionic.
Therefore, ionic compounds are a. KCl, c. BaCl2 and e. LiF.
Regards.
Answer:
pHe = 3.2 × 10⁻³ atm
pNe = 2.5 × 10⁻³ atm
P = 5.7 × 10⁻³ atm
Explanation:
Given data
Volume = 1.00 L
Temperature = 25°C + 273 = 298 K
mHe = 0.52 mg = 0.52 × 10⁻³ g
mNe = 2.05 mg = 2.05 × 10⁻³ g
The molar mass of He is 4.00 g/mol. The moles of He are:
0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol
We can find the partial pressure of He using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 3.2 × 10⁻³ atm
The molar mass of Ne is 20.18 g/mol. The moles of Ne are:
2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol
We can find the partial pressure of Ne using the ideal gas equation.
P × V = n × R × T
P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K
P = 2.5 × 10⁻³ atm
The total pressure is the sum of the partial pressures.
P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm
Answer:
There is an extra O2 molecule left over
Explanation:
