In a plastic sheet protector
<h3>
Answer:</h3>
0.387 J/g°C
<h3>
Explanation:</h3>
- To calculate the amount of heat absorbed or released by a substance we need to know its mass, change in temperature and its specific heat capacity.
- Then to get quantity of heat absorbed or lost we multiply mass by specific heat capacity and change in temperature.
- That is, Q = mcΔT
in our question we are given;
Mass of copper, m as 95.4 g
Initial temperature = 25 °C
Final temperature = 48 °C
Thus, change in temperature, ΔT = 23°C
Quantity of heat absorbed, Q as 849 J
We are required to calculate the specific heat capacity of copper
Rearranging the formula we get
c = Q ÷ mΔT
Therefore,
Specific heat capacity, c = 849 J ÷ (95.4 g × 23°C)
= 0.3869 J/g°C
= 0.387 J/g°C
Therefore, the specific heat capacity of copper is 0.387 J/g°C
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
