How many grams are in 1.11 moles of manganese sulfate, Mn3(SO4)7?
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<h3>Answer:</h3>
Temperature is 529.164 K
<h3>Explanation:</h3>We are given
Number of moles of Ne (n) = 0.019135 moles
Volume (V) = 878.3 mL
Pressure (P) = 0.946 atm
We are required to calculate the temperature;
We can do this using the ideal gas law equation which is;
PV = nRT, where P is the pressure, n is the number of moles, V is the volume, R is the ideal gas constant (0.082057 Latm/mol/K) and T is the temperature.
From the equation;
Therefore, the temperature will be 529.164 K.