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1000 mL=1L
25 mL = 0.025 L
125 mL = 0.125 L
M1V1=M2V2
0.15(0.125) = M2(0.025)
0.01875 = M2(0.025)
0.75 = M2
0.75 M
<u>Answer:</u> The temperature increase will be 31.70°C.
<u>Explanation:</u>
To calculate the increase in the temperature of the system, we use the equation:
where,
q = Heat absorbed = 36.5 kJ = 36500J
m = Mass of water = 275 g
c = Specific heat capacity of water = 
= change in temperature = ? °C
Putting values in above equation, we get:
Hence, the temperature increase will be 31.70°C.
Answer- 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
Given - Number of moles of Al(NO3)3 - 4 moles
Number of moles of NaCl - 9 moles
Find - Maximum amount of AlCl3 produced during the reaction.
Solution - The complete reaction is - Al(NO3)3 + 3NaCl --> 3NaNO3 + AlCl3
To find the maximum amount of AlCl3 produced during the reaction, we need to find the limiting reagent.
Mole ratio Al(NO3)3 - 4/1 - 4
Mole ratio NaCl - 9/3 - 3
Thus, NaCl is the limiting reagent in the reaction.
Now, 3 moles of NaCl produces 1 mole of AlCl3
9 moles of NaCl will produce - 1/3*9 - 3 moles.
Weight of AlCl3 - 3*133.34 - 400 grams
Thus, 400 grams of AlCl3 is the maximum amount of AlCl3 produced during the experiment.
