A chemistry student must write down in her lab notebook the concentration of a solution of sodium hydroxide. The concentration o
2 answers:
Answer:
The student should write the concentration as 9.72 x 10⁻² g/mL.
Explanation:
In this problem, the mass of sodium hydroxide is the difference between the initial mass of the cylinder and the mass of the cylinder after the addition of sodium hydroxide. It should be noted here that the figures should be rounded off according to the rules of significant figures. The mass is given as,
Here, only one figure is kept after the decimal. Among the two numbers whose difference is calculated, the number 1.5 contains one significant figure after its decimal. So, the answer is rounded off to one significant figure after the decimal.
The volume of the solution in the cylinder was 127.53 mL. So, the concentration of NaOH will be:
According to the rules of significant figures for division, the answer should be rounded off to the least number of significant figures in the problem. Here, it is three significant figure due to the number 12.4 in the numerator.
You might be interested in
Answer:
A) .
B) .
C) 0.9 mol.
D) Increasing both temperature and pressure.
Explanation:
Hello,
In this case, given the information, we proceed as follows:
A)
B) For the calculation of Kc, we rate the equilibrium expression:
Next, since at equilibrium the concentration of ammonia is 0.6 M (0.9 mol in 1.5 dm³ or L), in terms of the reaction extent , we have:
Next, the concentrations of nitrogen and hydrogen at equilibrium are:
Therefore, the equilibrium constant is:
C) In this case, the equilibrium yield of ammonia is clearly 0.9 mol since is the yielded amount once equilibrium is established.
D) Here, since the reaction is endothermic (positive enthalpy change), one way to increase the yield of ammonia is increasing the temperature since heat is reactant for endothermic reactions. Moreover, since this reaction has less moles at the products, another way to increase the yield is increasing the pressure since when pressure is increased the side with fewer moles is favored.
Best regards.