Tn 4351 was originally isolated from bacteroides fragilis [30] . The transposon was successfully introduced into Cytophaga succinicans, Flavobacterium meningosepticum, Flexibacter canadiansis, Flexibacter strain SFI and Sporocytophaga myxococcoides by conjugation [25]. Tn 4351carries two antibiotic resistance gene. One of the codes for resistance to erythromycin and clindamycin which is expressed in bactroides but not in E.Coli. The other gene codes for resistance in tetracycline and is expressed in aerobically grpwn E. coli, but not in anaerobically grpwn E. coli or in bacteroides.
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
I organized four different tests; pH paper, alkalinity tests, the number of rocks neutralizing acid, and the number of rocks that don’t neutralize acid. The average for pH paper was seven. The pH is the numeric scale used to specify the acidity or alkalinity of an aqueous solution. The pH scale measures acidity and alkalinity. the pH scale goes from zero, which is an acid reading, to fourteen, which is an alkaline reading.
A solution with a pH of 7.0 is considered _______. a. Acidic c. alkaline (basic) b. Neutral d. saturated 9. If the concentration of the H+ ion is greater than the concentration of the OH- ion, the solution is said to be ________. a. Acidic c. alkaline (basic) b. Neutral d. saturated 10.
The unknown compound was first reacted with an acid. To begin, 0.50 grams of KCl was mixed with 5 mL of water. Then, 1 mL of 6 M H2SO4 was added to the solution. Secondly, the unknown compound was reacted with a base. Exactly 0.50 grams of KCl was mixed with 5 mL of water, and 1 mL of 1 M NaOH was added to the solution next.
Research Question: To investigate and compare how different temperature (5℃, 15℃, 25℃, 35℃, 45℃) can affect the concentration of carbon dioxide in soda water through titration with sodium hydroxide solution. Introduction: Carbon dioxide plays an important role in soft drinks. Soda water is manufactured by pumping carbon dioxide into water under high pressure. Carbon dioxide dissolves in water to form carbonic acid, which is the fizz we find in soft drinks. CO2 + H2O ⇌
Joshua Miller 12/18/17 Fermentation Lab report Introduction The term fermentation refers to the chemical breakdown of a substance by bacteria, yeasts, or other microorganisms, typically involving effervescence and the giving off of heat (wikipedia). Sugars are converted to ethyl alcohol when fermentation happens. In this experiment we determined if yeast cells undergo fermentation when placed in a closed flask with no oxygen. Glucose and yeast are mixed together in a closed flask and allowed to incubate for about one hour.
Acids are proton donors in chemical reactions which increase the number of hydrogen ions in a solution while bases are proton acceptors in reactions which reduce the number of hydrogen ions in a solution. Therefore, an acidic solution has more hydrogen ions than a basic solution; and basic solution has more hydroxide ions than an acidic solution. Acid substances taste sour. They have a pH lower than 7 and turns blue litmus paper into red. Meanwhile, bases are slippery and taste bitter.
The chemical equation for this experiment is hydrochloric acid + sodium thiosulphate + deionised water (ranging from 25ml to 0ml in 5ml intervals) sodium chloride + deionised water (ranging from 25ml to 0ml in 5ml intervals) + sulphur dioxide + sulphur. As a scientific equation, this would be written out as, NA2S2O3 + 2HCL + H2O (ranging from 25ml to 0ml in
The equation of the reaction between sodium hydroxide and ethanoic acid is as follows: CH3COOH + NaOH → CH3COONa + H2O We can measure the end point of titration process and we can also measure the amount of reactants. The concentration of ethanoic acid in the vinegar can be determined through stoichiometric calculations, Using the values obtained from the titration, and also the chemical equation as a reference. Phenolphthalein indicator is used in this acid-base titration Equipment and materials:
When carbohydrate is utilized, acids are formed which changes the colour of the medium from green to yellow
Aim: To find out the relationship between the greater concentration of sodium thiosulfate when mixed with hydrochloric acid and the time it takes for the reaction (the time it takes for the solution to turn cloudy) to take place and to show the effect on the rate of reaction when the concentration of one of the reactants change. Introduction: The theory of this experiment is that sodium thiosulfate and hydrochloric acid reach together to produce sulfur as one of its products. Sulfur is a yellow precipitate so, the solution will turn to yellow color while the reaction is occurring and it will continue until it will slowly turn completely opaque. The reaction of the experiment happens with this formula: “Na2 S2 O3 + HCL =
In direct titrations, the number of moles of acid can be easily derived by simply manipulating with the values of acid and base given in the experiment. In back titration, excessive volumes of acid are always added. Of which, only a certain quantity would be neutralised. The number of moles of acid is eventually derived from titrating this excess acid with a strong base and using mole fractions to calculate. The quantity of acid neutralised is obtained by subtracting the moles of acid given at the start of the experiment, with the moles of acid titrated.
The mixture were stirred by using a glass rod until the mixture is fully dissolved. The solution were tested by using calibrated pH meter to get the pH value of the solution. Results and Discussions pH ratio between acid and base: 7.3 = 6.82 + x x = 0.48 0.48 = log ([base])/([acid]) 100.48 =base/acid salt/acid = 3.02 There, 1 acid : 3 base calculate number of mole of acid and base to find the mass : molar = mol/L 50 mM = (mol )/(0.5 L) mol = 25 mol number of mole of NaH2PO4 25/4 = acid = 6.25 mol number of mole of Na2HPO4 25/4 × 3 = salt = 18.75 mol to calculate the mass of the acid and base : Mass of NaH2PO4 (6.25 mol)/(119.98 g/mol)=0.052g
That caused a new initial reading of NaOH on the burette (see Table1 & 2). The drops were caused because the burette was not tightened enough at the bottom to avoid it from being hard to release the basic solution for titrating the acid. The volume of the acid used for each titration was 25ml. The volume of the solution was then calculated by subtracting the initial volume from the final volume. We then calculated the average volume at each temperature.
