I. Introduction This experiment uses calorimetry to measure the specific heat of a metal. Calorimetry is used to observe and measure heat flow between two substances. The heat flow is measured as it travels from a higher temperature to a lower one. Specific heat is an amount of heat required to raise the temperature of one gram of anything one degree Celsius.
In the sodium iodide test, the alkyl halide is added to sodium iodide in acetone. In this test, primary halides precipitate the fastest while secondary halides need to be heated in order for a reaction to occur. Comparison of the rates of precipitation of the obtained product to standard 1° and 2° bromide solutions will show whether the product is a primary or secondary
Given that 3.327 g of product was produced during the reaction, how many grams of water were released as water vapor? (moles = mass/molecular weight). From observation it appeared 1.673 grams of water were released as water vapor. 5. Write a balanced equation for the decomposition of copper carbonate hydroxide hydrate.
In addition, phenolphthalein was added as an indicator. The aliquots were titrated against sodium hydroxide (NaOH) solution until end point was reached, after which volume of NaOH consumed was recorded. The value of the rate constant, k, obtained was 0.0002 s-1. The experiment was then repeated with 40/60 V/V isopropanol/water mixture and a larger value of k = 0.0007 s-1 was obtained. We concluded that the rate of hydrolysis of (CH3)3CCl is directly proportional to water content in the solvent mixture.
10, is a linear curve for 4-NP reduction using AuNPs. It was observed that the increase in temperature helps the rate of reaction to increase. The activation energy was calculated from the slope of the straight line and was found to be 7.4 ± 1.34 k Cal/mol. The above results are of clear indication that catalysis usually takes place on the surface of the nanoparticles. 3.8 Catalytic reduction of potassium hexacyanoferrate (III) The electron transfer reaction between hexacyanoferrate (III) and sodium borohydride results in the formation of hexacyanoferrate (II) ion and dihydrogen borate ion and this reaction is strongly catalyzed by AuNPs.
For example, If 2.24L of oxygen gas at STP has given, then it can be easily conclude that the volume would contain one tenth of the mole of oxygen gas. That is, total number of 6.022 x 1022 molecules of oxygen will be present. CONSEQUENCES OF AVOGADRO'S LAW There are a few important consequences of Avogadro's law The molar volume of all ideal gases at 1 atm pressure and 0°C is 22.4 L the volume increases. the amount of gas increases, only If temperature and pressure of a gas are constant when the volume decreases and amount of gas decreases, If pressure and temperature of a gas are constant. The fallowing graph shows the relationship between mass (n) and volume (v) as shown in Avogadro's Law.
This value makes sense since all of the values for viscosity of Mystery 1 are between those collected for the 10 wt% sucrose and 20 wt% sucrose solutions. It’s average density of 1.037 g/mL is near that of the 10 wt% sucrose solution’s of 1.038 g/mL, showing it is closer to 10 wt% sucrose than 20 wt% sucrose. The concentration of Mystery 2, calculated using average viscosities collected at each temperature, and then an overall average of the concentrations calculated from Equations 3-5, was found to be 28.725 wt% sucrose. This value makes sense since most of the values for viscosity of Mystery 2 are between those collected for the 20 wt% sucrose and 30 wt% sucrose solutions. It’s average density of 1.0975 g/mL is between that of the 20 wt% sucrose solution’s of 1.079 g/mL and that of the 30 wt% sucrose solution’s of 1.129 g/mL.
The solution was then titrated with AgNO3 and volume used to generate the colour change was recorded. From the volume of AgNO3 used for the titration, the moles of Ag+ were calculated and found to be 0.005 moles of Ag+. Silver ions and iodine ions react in mole ratio of 1:1 to form the precipitate AgI (s). From the mole ratio it was calculated that 0.005 moles of I- reacted and thus 100% of I- was
To investigate the effect of agitation time and feed concentration on percentage removal of phenol batch experiments were carried out at room temperature in the range of 100 to 700 mg/l of initial feed concentration and the results are shown in Fig. 2.Experimental conditions used were agitation speed of 200rpm, biosorbent dosage 0.15 g and pH 7.From figure it is observed that with increase in agitation time from 0 to 2 days, the percentage removal increased from 8% to 98%.Phenol adsorption from bulk liquid to biosorbent is high when the agitation time was increased. This leads to a higher adsorption of phenol on the biosorbent surface and it could reach an equilibrium beyond which percentage removal may not increase. In the present study equilibrium is reached at 4 days. The initial phenol concentration provides an important driving force to overcome all mass transfer limitations of phenol between the aqueous and sorbent phases.
What is the mass of a 22.3 cm3 sample? Let’s rearrange the density equation to solve for mass: d= mass volume ⇒d×volume=mass⇒22.59 g c m 3 ×22.3 c m 3 =504g d=massvolume⇒d×volume=mass⇒22.59gcm3×22.3cm3=504g By rearranging the equation, cm3 gets cancelled out, leaving g as a result. Quick Check: 2.8.b What is the volume of a 5.98 g sample of sodium chloride? Sodium chloride has a density of 2.16 g/cm3. 3.00 g of a sample was placed in a graduated cylinder with 10.0 mL of water.