Formula for Molar mass is: MM = grams/moles
First step is to find moles. The formula to find it is:
PV = nRT (Odde, 2015)
Convert all the units into standard units.
P = 777 torr = 1.02 atm ; V = 125mL = 0.125L ; R = 0.082 K-1 L mol-1 atm; T = 126 C = 399.15 K
n = PV/RT = 0.00389276 mol
Since there are 0.281 grams, divide it with the number of moles to get molar mass
Molar Mass = grams/mol = 0.281/0.00389276 = 72 g/mol
In order to find the volume of H2. First step is to find the number of moles of H20. By using the formula n = m/M;
In order to find number of moles, we will divide the mass, which is 15, by the molar mass of H20, which is 18 (1×2 + 2×8).
m = 15 and M = 18.
n = m/M = 15/18 = 0.8333.
After finding moles of H2 which is 0.8333. Use the equation PV = nRT.
Convert all units into standard units.
P = 745 torr = 0.98 atm ; T = 20 C = 293 K ; R = 0.082 L atm K-1 mol-1
V = nRT/P = (0.833)*(0.0821)*(293) / 0.98 = 2.011 L
15g of H2O will produce 2.011 L of H2.
Lab 6 Assignment
Titration is a process in which a titrant, which is a solution whose concentration is known, is added very slowly into a solution whose concentration is not known. This is done until the change of color takes place, which is also called neutralization. It is a method to find out how much solute is to be added in order for the reaction to happen.
In the video, the acid base titration has been demonstrated. CH3COOH is an acid and NaOH is the base and the process of titration is done between them.
NaOH + CH3COOH = H2O + Na(CH3COO)
The process shows acidic solution has CH3COO– and H3O+ ions whereas basic solution has Na+ and OH– ions and when the acidic and basic solution combines, then these two opposite ions i.e. negative and positive ions are bound together with each other.
The reactants used are acetic acid and sodium hydroxide.
The glassware is a burette, which helps to take the measurement of the liquid more precisely and accurately. Sodium hydroxide was measured through it.
The first step is to balance the equation. Then a known concentration and volume of acid should be taken. Now in order to get the moles of the acid, through molarity and acid volume, moles are to be found. As the equation is balanced, next step is to determine the ratio of moles. The equation we have has equal one to one. Moles of base in this balanced equation will be found, which were used in the reaction. For Example, 150mL of 0.100 moles CH3COOH to moles of NaOH. (150mL CH3COOH) x (0.100 moles CH3COOH/L CH3COOH) x (1L CH3COOH/1000mL CH3COOH) x (1 mole CH3COOH/ 1 mole NaOH) = 0.15 NaOH moles.
The reaction between NaOH and CH3COOH is done with the ratio of 1:1. Equal moles are reacted. The concentration of CH3COOH in vinegar can be found out by dividing the moles by the vinegar volume. For instance, vinegar volume is 20mL. Then we will get 0.89M CH3COOH. 1000mL/1L x 0.01778mol x 1/20mL = 0.89M.
The cause of the solution to be basic is because of the presence of OH– and the cause of it to be acidic is due to H+. The reaction is acid basic in nature and so the reaction gives water and salt.
The reaction is acid basic in nature and so the reaction gives water and salt.
The pH of an acid can be found out. The fusion of two 2 chemical substance with analyte and focus. Burette is used for the titration purpose. The moles of analyte and titrant are equal from the equality focus. In a given test, the substance response’s purpose in the process of titration is the titrant’s focus which is equal in stoichiometric sense to the analyte’s moles.
pH = -log(H3O+)
= -log(8.9 x 10-1)
The demonstration of Dr. Yee showed that sodium hydroxide should be used to clean titration. One more thing that was demonstrated is that the valve turned on and then off by the stopcock as soon as the addition of liquid took place. Dr. Yee ad also clear the tip if it was not cleared, in order for the flow to be constant.
The reaction between hydrochloric acid and limestone is:
CaCO3 + 2HCl = CO2 + H2O + Ca2
When a limestone reacts with acid, it will give off gas bubbles. That is the first thing I will be looking for. If the reaction produces gas bubbles, it will mean that it is a reaction between carbonates(mineral or rock) and an acid in it.
C(NaOH) x V(base) = C(HCl) x V(acid)
C(HCl) = (39.25mL) (0.1522mol) / 35mL
= 0.170 mol HCl
Q16) As soon as the color of the liquid changes to pink, this is the indication of the completion of the reaction.
The name used by Dr. Yee is Phenolphthalein.
The name of the indicator is Phenolphthalein and the way it is used is by adding it into the liquid and when the liquid changes its color to pink, then stop adding it.
Student # 1 Stephen
The first is done very accurately and effectively. The key to solve a question is first to find the correct answer and another more effective way to do the question is to find the simplest way to answer any problem, with the minimum steps. The question 1 is done with minimum possible steps. The answer is correct i.e. 12.1L. But the second question is incomplete. The way to go about that question is first write the balanced equation of the reaction. The next step is to find the oxygen’s moles in the product part of the reaction (O2). Then you have to make sure all the units of pressure are same I=Convert all of them into atm. Then the final step is to use the gas law formula which is, PV = nRT. Rearrange the equation to find the volume. V = nRT/P. You will get your answer.
Student # 2 Steigwalker
The best way to solve a question is to first define all the known and unknown variable mentioned in the question. For instance, in the first question, the volume is to be found out. Mention what V2 represent and similarly define what other variables mean. This should be done by giving the heading of “Data” and then give heading of what it is there to find, which is V2. Then give the heading to solution, and start solving it. This will give a neat look to your answer and it will be easy for the teacher to check the answer. Also highlight the answers. The answers are correct but it would be better if every step is explained in one sentence or more so that the teacher would know you have done it by yourself.
Odde, D. J. (2015). Mitosis, diffusible crosslinkers, and the ideal gas law. Cell, 160(6), 1041–1043.