The answer is Cs and F. S and O are both non-metals and if I am not mistaken they are both on the right side of the periodic table. S and O exhibit a covalent bond, the other pairs do too except for the first pair of atoms. Therefore, the answer is A. Cs and F.
Answer:
a. 5.36x10⁻⁴ g/mL
b. 4.29x10⁻⁵ g/mL
Explanation:
As the units for concentration are not specified, I'll respond using g/mL.
a. We <em>divide the sample mass by the final volume</em> in order to <u>calculate the concentration</u>:
- 0.268 g / 500 mL = 5.36x10⁻⁴ g/mL
b. We can use C₁V₁=C₂V₂ for this question:
- 8.00 mL * 5.36x10⁻⁴ g/mL = C₂ * 100.00 mL
Answer:
41.17g
Explanation:
We are given the following parameters for Flourine gas(F2).
Volume = 5.00L
Pressure = 4.00× 10³mmHG
Temperature =23°c
The formula we would be applying is Ideal gas law
PV = nRT
Step 1
We find the number of moles of Flourine gas present.
T = 23°C
Converting to Kelvin
= °C + 273k
= 23°C + 273k
= 296k
V = Volume = 5.00L
R = 0.08206L.atm/mol.K
P = Pressure (in atm)
In the question, the pressure is given as 4.00 × 10³mmHg
Converting to atm(atmosphere)
1 mmHg = 0.00131579atm
4.00 × 10³ =
Cross Multiply
4.00 × 10³ × 0.00131579atm
= 5.263159 atm
The formula for number of moles =
n = PV/RT
n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K
n = 1.0834112811moles
Step 2
We calculate the mass of Flourine gas
The molar mass of Flourine gas =
F2 = 19 × 2
= 38 g/mol
Mass of Flourine gas = Molar mass of Flourine gas × No of moles
Mass = 38g/mol × 1.0834112811moles
41.169628682grams
Approximately = 41.17 grams.
Answer:
2.77 mL of boiling water is the minimum amount which will dissolve 500 mg of phthalic acid.
Explanation:
We know from the problem that 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C.
Now we devise the following reasoning:
If 18 g of phthalic acid are dissolved in 100 mL of water at 99 °C
Then 0.5 g of phthalic acid are dissolved in X mL of water at 99 °C
X = (0.5 × 100) / 18 = 2.77 mL of water
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