Phosphorus + Sulfur ------> Phosphorus sulfide
2P + 3S ------> P2S3
Hope it helped!
Answer:
10043.225 J
Explanation:
We'll begin by calculating the amount of heat needed to change ice to water since water at 0°C is ice. This is illustrated below:
Mass (m) = 15.5g
Latent heat of fussion of water (L) = 334J/g
Heat (Q1) =..?
Q1 = mL
Q1 = 15.5 x 334
Q1 = 5177 J
Next, we shall calculate the amount of heat needed to raise the temperature of water from 0°C to 75°C.
This is illustrated below:
Mass = 15.5g
Initial temperature (T1) = 0°C
Final temperature (T2) = 75°C
Change in temperature (ΔT) = T2 – T1 = 75 – 0 = 75°C
Specific heat capacity (C) of water = 4.186J/g°C
Heat (Q2) =?
Q2 = MCΔT
Q2 = 15.5 x 4.186 x 75
Q2 = 4866.225 J
The overall heat energy needed is given by:
QT = Q1 + Q2
QT = 5177 + 4866.225
QT = 10043.225 J
Therefore, the amount of energy required is 10043.225 J
NO, it should not be the process is still the same . the only factors about it that should change is the experiment itself.:)
Answer:
the concentration of bicarbonate is <em>[HCO₃⁻] = 0,03996 M </em>and carbonate is <em>[CO₃²⁻] = 3,56x10⁻⁵ M.</em>
Explanation:
Carbonate-bicarbonate is:
HCO₃⁻ ⇄ CO₃²⁻ + H⁺ With pka = 10,25
Using Henderson-Hasselbalach formula:
pH = pka + log₁₀![\frac{[CO_{3}^{2-}]}{[HCO_{3}^-]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO_%7B3%7D%5E%7B2-%7D%5D%7D%7B%5BHCO_%7B3%7D%5E-%5D%7D)
7,2 = 10,25 + log₁₀
8,91x10⁻⁴ = <em>(1)</em>
Also:
0,040 M = [CO₃²⁻] + [HCO₃⁻] <em>(2)</em>
Replacing (2) in 1:
<em>[HCO₃⁻] = 0,03996 M</em>
Thus:
<em>[CO₃²⁻] = 3,56x10⁻⁵ M</em>
I hope it helps.
