61.9 grams
2.33x10^23 molecules (159.7 g Fe2O3/6.02x10^23 molecules) = 61.9 g
Answer:
Molar heat enthalpy of KBr = -19.89 kJ/mol
Explanation:
Change in temperature (Δt) = 0.370 K
Heat capacity = 2.71 kJ ⋅ K^-1
Heat absorbed by calorimeter = heat capacity × change in temperature
= 2.71× 0.370
= 1.0027 kJ
Molar mass of KBr = 119 g/mol
No. of moles of KBr = 6.00/119
= 0.0504 mol
Heat absorbed by the calorimeter is given by KBr.
Now calculate the heat released by per mol of KBr as follows:
Heat released by per mol of KBr = 1.0027 kJ / 0.0504 mol
=19.89 kJ/mol
Heat is released therefore, sign will be negative.
Molar heat enthalpy of KBr = -19.89 kJ/mol
Answer:
hi basically also also Aussie
When titrating a strong monoprotic acid and KOH at 25 degrees Celcius, the pH will be approximately equal to 7.00. Since a strong acid and a strong base reacts through a neutralization reaction. The pH of a neutral solution is 7.00. <span />
Hello, my name is Zalgo and I am going to be providing you with the scientific knowledge that you seem to need. In the year 1848, William Thomson or also know as Lord Kelvin was the founder of Zero Kelvin or most commonly known as "Absolute Zero". It is the point to where something has completely frozen. Like other temperature scales, the boiling and freezing points of water are important factors in establishing the scale's range. There are 100 degrees between in which the water boils or freezes (Boils --> 373.16 K) (Freeze --> 273.16)
I hope this helps.
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