The correct answer that would best complete the given statement above would be the term STEEL. The Bessemer process created strong, lightweight called steel. <span>The Bessemer process is the first process of making or mass producing steel. Hope this is the answer that you are looking for. Thanks for posting!</span>
To balance this equation, first we should consider balancing C because it only presents in one reactant and one product. Assuming the coefficient of C6H6 is 1, there are 6 C's in the reactant, so it generates 6CO2. Then consider balancing H for the same reason. If the coefficient of C6H6 is 1, there are 6 H's in the reactant, so it generates 3H2O.
Now that the coefficient of the products are determined, we can balance O. There are 6*2=12 O's in CO2 and 3*1=3 O's in H2O. So the total number of O in the products is 12+3 = 15. O2 is the only reactant that contains O, so to balance the equation, the coefficient of O2 should be 15/2.
Now the equation looks like:
C6H6 + 15/2O2 ⇒ 6CO2 + 3H2O.
Times both sides of the equation by 2 results the final answer:
2C6H6 + 15O2 ⇒ 12CO2 + 6H2O
First, we have to remember the molarity formula:
Part 1:
In this case, our solute is sodium nitrate (NaNO3), and we have the mass dissolved in water, then we have to convert grams to moles. For that, we need the molecular weight:
Then, we calculate the moles present in the solution:
Now, we have the necessary data to calculate the molarity (with the solution volume of 200 mL):
The molarity of this solution equals 0.2339 M.
Part 2:
In this case, we have the same amount (in moles and mass) of sodium nitrate, but a different volume of solution, then we only have to change it:
So, the molarity of this solution is 0.1701 M.
<span>The liquid form of matter is usually more dense than its gas form. This is because liquid molecules are closer together compared to gas molecules. An exception, however, is water. Water's solid form or ice is less dense than its liquid form because of the orientation of hydrogen bonds that lowers its density.</span>
