From moles to grams C4H10 weighs 174.3666grams
The question is incomplete, here is the complete question:
A chemist prepares a solution of iron (III) bromide
by measuring out 2.78 g of iron (III) bromide into a 50. mL volumetric flask and filling the flask to the mark with water.
Calculate the concentration in mmol/L of the chemists iron (III) bromide solution. Be sure your answer has the correct number of significant digits.
<u>Answer:</u> The concentration of iron(III) bromide solution is 0.19 M
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

We are given:
Given mass of iron(III) bromide = 2.78 g
Molar mass of iron(III) bromide = 298.6 g/mol
Volume of solution = 50. mL
Putting values in above equation, we get:

Hence, the concentration of iron(III) bromide solution is 0.19 M
Answer:
52.79 dm3 of SO2 will be produced when 975kJ/mol of heat are liberated.
Explanation:
In the reaction given;
2PbS(s) + 3O2(g) 2PbO(s) + 2SO2(g) H = -827.4 kJ/mol
-827.4 kJ/mol of heat liberates 2 moles of SO2 in the reaction involving lead and oxygen
At STP, -827.4 kJ/mol of heat liberate 22.4 * 2 dm3 of SO2
So therefore, 975 kJ/mol of heat will liberate
= 975 * 22.4 * 2 / -827.4
= 43 680 / -827.4
= 52.79 dm3 of SO2.
52.79 dm3 of SO2 will be produced at STP if 975 kJ/mol of heat are liberated.
That would be choice A.
Protons and Neutrons.