Answer:
B)
Explanation:
This should be the correct answer, lmk if it's not
The equilibrium will shift to the left or the backward reaction since addition of <span>CH3COONa will add more CH3COO- ions to the solution. The formation of reactants are promoted.</span>
Answer:
![T_2=261.46\ K](https://tex.z-dn.net/?f=T_2%3D261.46%5C%20K)
Explanation:
It is given that,
Original temperature, ![T_1=323^{\circ}C=596.15\ K](https://tex.z-dn.net/?f=T_1%3D323%5E%7B%5Ccirc%7DC%3D596.15%5C%20K)
Original volume, ![V_1=2.85\ L](https://tex.z-dn.net/?f=V_1%3D2.85%5C%20L)
We need to find the temperature if the volume of the balloon to be shrink to 1.25 L.
According to Charles law, at constant pressure, ![V\propto T](https://tex.z-dn.net/?f=V%5Cpropto%20T)
It would means, ![\dfrac{V_1}{V_2}=\dfrac{T_1}{T_2}](https://tex.z-dn.net/?f=%5Cdfrac%7BV_1%7D%7BV_2%7D%3D%5Cdfrac%7BT_1%7D%7BT_2%7D)
T₂ = ?
![T_2=\dfrac{V_2T_1}{V_1}\\\\T_2=\dfrac{1.25\times 596.15}{2.85}\\\\T_2=261.46\ K](https://tex.z-dn.net/?f=T_2%3D%5Cdfrac%7BV_2T_1%7D%7BV_1%7D%5C%5C%5C%5CT_2%3D%5Cdfrac%7B1.25%5Ctimes%20596.15%7D%7B2.85%7D%5C%5C%5C%5CT_2%3D261.46%5C%20K)
So, the new temperature is 261.46 K.
<em>M CaCl₂: 40+(35,5×2) = 111 g/mol</em>
6,02·10²³ molecules ---------- 111g
X molecules --------------------- 75,9g
X = (75,9×<span>6,02·10²³)/111
X = <u>4,116</u></span><span><u>·10²³</u> molecules of CaCl</span>₂
:)