The number of energy levels increases as you move down a group as the number of electrons increases. Each subsequent energy level is further from the nucleus than the last.
Answer:
Writing Formulas of Ionic Compounds
The cation is written first, followed by the monatomic or polyatomic anion. The subscripts in the formula must produce an electrically neutral formula unit.
Answer:
874.89 mmHg
Explanation:
Using the combined gas law equation as follows:
P1V1/T1 = P2V2/T2
Where;
P1 = initial pressure (mmHg)
P2 = final pressure (mmHg)
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
Based on the information provided in this question;
P1 = 950 mm Hg
P2 = ?
V1 = 25 L
V2 = 30 L
T1 = 155°C = 155 + 273 = 428K
T2 = 200°C = 200 + 273 = 473K
Using P1V1/T1 = P2V2/T2
950 × 25/428 = P2 × 30/473
23750/428 = 30P2/473
55.49 = 0.063P2
P2 = 55.49 ÷ 0.063
P2 = 874.89 mmHg
Answer:
0.15 M
Explanation:
14.6/46.07g/mol = 0.316 mol Ethanol.
0.316/2.1= 0.15 M
Answer:
0.001 mole of NaF.
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 100 mL
Molarity = 0.01 M
Mole of NaF =?
Next, we shall convert 100 mL to litre (L). This can be obtained as follow:
1000 mL = 1 L
Therefore,
100 mL = 100 mL × 1 L / 1000 ml
100 mL = 0.1 L
Thus, 100 mL is equivalent to 0.1 L.
Finally, we shall determine the number of mole of NaF present in the solution. This can be obtained as follow:
Volume of solution = 0.1 L
Molarity = 0.01 M
Mole of NaF =?
Molarity =mole /Volume
0.01 = mole of NaF / 0.1
Cross multiply
Mole of NaF = 0.01 × 0.1
Mole of NaF = 0.001 mole.
Thus, 0.001 mole of NaF is present in 100 mL of the solution.
