Answer:
(b) fully filled valence s orbitals
Explanation:
Electron configuration of Be: 1s22s2
2s2 is fully filled
Answer:
2.2 °C/m
Explanation:
It seems the question is incomplete. However, this problem has been found in a web search, with values as follow:
" A certain substance X melts at a temperature of -9.9 °C. But if a 350 g sample of X is prepared with 31.8 g of urea (CH₄N₂O) dissolved in it, the sample is found to have a melting point of -13.2°C instead. Calculate the molal freezing point depression constant of X. Round your answer to 2 significant digits. "
So we use the formula for <em>freezing point depression</em>:
In this case, ΔTf = 13.2 - 9.9 = 3.3°C
m is the molality (moles solute/kg solvent)
- 350 g X ⇒ 350/1000 = 0.35 kg X
- 31.8 g Urea ÷ 60 g/mol = 0.53 mol Urea
Molality = 0.53 / 0.35 = 1.51 m
So now we have all the required data to <u>solve for Kf</u>:
Answer:
0.1 M
<h3>
Explanation:</h3>
- Molarity refers to the concentration of a solution in moles per liter.
- It is calculated by dividing the number of moles of solute by the volume of solvent;
- Molarity = Moles of the solute ÷ Volume of the solvent
<u>In this case, we are given;</u>
- Number of moles of the solute, NH₄Cl as 0.42 moles
- Volume of the solvent, water as 4200 mL or 4.2 L
Therefore;
Molarity = 0.42 moles ÷ 4.2 L
= 0.1 mol/L or 0.1 M
Thus, the molarity of the solution will be 0.1 M
Answer:
increase
Explanation:
Let's suppose we have a sample of air in a closed container. We heat the container and we want to predict what would happen to the pressure.
According to Gay-Lussac's law, the pressure of a gas is directly proportional to its absolute temperature.
Thus, if we increased the temperature of the air by heating it, its pressure would increase.
If a sample of air in a closed container was heated, the total pressure of the air would increase.
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