Answer:
I know that the 100-mL graduated cylinders are always read to 1 decimal place.
I think for 50 mL graduated cylinders, it lets you measure volumes up to 50.0 mL to the nearest 0.1 or 0.2 mL, depending on your exact cylinder.
Answer:
10 kg of ice will require more energy than the released when 1 kg of water is frozen because the heat of phase transition increases as the mass increases.
Explanation:
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In this case, since the melting phase transition occurs when the solid goes to liquid and the freezing one when the liquid goes to solid, we can infer that melting is a process which requires energy to separate the molecules and freezing is a process that releases energy to gather the molecules.
Moreover, since the required energy to melt 1 g of ice is 334 J and the released energy when 1 g of water is frozen to ice is the same 334 J, if we want to melt 10 kg of ice, a higher amount of energy well be required in comparison to the released energy when 1 kg of water freezes, which is about 334000 J for the melting of those 10 kg of ice and only 334 J for the freezing of that 1 kg of water.
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Answer:
CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
Explanation:
To answer this question we must know Kb of CH3CH2NH2 is 5.6x10⁻⁴, and for C6H5NH2 is 4.0x10⁻¹⁰. And the CH3CH2NH3+ and C6H5NH3+ are related with these substances because are their conjugate base. That means:
pKa of CH3CH2NH3+ = CH3CH2NH2; C6H5NH3+ = C6H5NH2
Also, Kw / Kb = Ka
Thus:
pKa of CH3CH2NH3+/CH3CH2NH2 is:
Kw / kb = Ka = 1.79x10⁻¹¹
-log Ka = pKa
pKa = 10.75
pKa of C6H5NH3+/ C6H5NH2 is:
Kw / kb = Ka = 2.5x10⁻⁵
-log Ka = pKa
pKa = 4.6
That means CH3CH2NH3+/CH3CH2NH2 would have the largest pKa
The Earth's crust or is it the continental plates one of the two but they push against each other costing a tsunami.
Question:
<span>A sample of nitrogen gas had a volume of 500mL, a pressure in its closed container of 740 torr and a temperature of 25°c. what was the volume of gas when the temperature was changed to 50°c and the new pressure was 760 torr?
Answer:
Data Given:
V</span>₁ = 500 mL
P₁ = 740 torr
T₁ = 25 °C + 273 = 298 K
V₂ = ?
P₂ = 760 torr
T₂ = 50 °C + 273 = 323 K
Solution:
Let suppose the gas is acting Ideally, then According to Ideal Gas Equation,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = (P₁ V₁ T₂) ÷ (T₁ P₂)
Putting Values,
V₂ = (740 torr × 500 mL × 323 K) ÷ (298 K × 760 torr)
V₂ = 527.68 mL
