Your answer would be 58.12g/mol ;)
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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:
Explanation:
Hello,
In this case, for a concentration of 0.42 M of benzoic acid whose Ka is 6.3x10⁻⁵ in 0.33 M sodium benzoate, we use the Henderson-Hasselbach equation to compute the required pH:
![pH=pKa+log(\frac{[base]}{[acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29)
Whereas the concentration of the base is 0.33 M and the concentration of the acid is 0.42 M, thereby, we obtain:
![pH=-log(Ka)+log(\frac{[base]}{[acid]} )\\\\pH=-log(6.3x10^{-5})+log(\frac{0.33M}{0.42M} )\\\\pH=4.1](https://tex.z-dn.net/?f=pH%3D-log%28Ka%29%2Blog%28%5Cfrac%7B%5Bbase%5D%7D%7B%5Bacid%5D%7D%20%29%5C%5C%5C%5CpH%3D-log%286.3x10%5E%7B-5%7D%29%2Blog%28%5Cfrac%7B0.33M%7D%7B0.42M%7D%20%29%5C%5C%5C%5CpH%3D4.1)
Regards.
Answer:
it make poisionus police come and took away
250 kJ/87.9 KJ per mole Cl2 * 71g/mole= 202 g It is D for plato users
