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The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.
The specific heat has been defined as the amount of heat required to raise the temperature of 1 gram of substance by 1 degree Celsius.
The specific heat has been expressed as:
The iron and calorimeter are in side the closed system. Thus, the energy released by the sample, has been equivalent to the energy absorbed by the calorimeter.
The given mass of calorimeter has been,
The specific heat of the calorimeter has been,
The change in temperature of the calorimeter has been,
Substituting the values for heat released:
The amount of heat released by the sample has been 22.54 kJ. Thus, option C is correct.
Learn more about specific heat, here:
Hello!
To solve this problem we'll use the Henderson-Hasselbach equation, but first we need the vale for the pKa of Benzoic acid, which is pKa= -log(Ka)=4,19
Now, we apply the equation as follows:
![pH=pKa + log ( \frac{[C_6H_5COONa]}{[C_6H_5COOH]} )=4,19+log( \frac{0,15M}{0,25M} )=3,97](https://tex.z-dn.net/?f=pH%3DpKa%20%2B%20log%20%28%20%5Cfrac%7B%5BC_6H_5COONa%5D%7D%7B%5BC_6H_5COOH%5D%7D%20%29%3D4%2C19%2Blog%28%20%5Cfrac%7B0%2C15M%7D%7B0%2C25M%7D%20%29%3D3%2C97%20)
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is 3,97
Have a nice day!
To solve this problem we'll use the Henderson-Hasselbach equation, but first we need the vale for the pKa of Benzoic acid, which is pKa= -log(Ka)=4,19
Now, we apply the equation as follows:
So, the pH of this solution of Sodium Benzoate and Benzoic Acid is 3,97
Have a nice day!
Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V
