Kushal+Jain

 = = ** Now Brinjal are dangerous for life…………… ** You would be surprise and also been shocked to hear that now Brinjal’s are dangerous for us. A small vegetable can take a life of yours. This is the problem = = = =   which India is facing. This is the current issue and also a hot topic in India. And every one is worry about the new type of Brinjal which is BT-Brinjal. From government to normal person is thinki = = = =  ng on BT-Brinjal. For this kind of species many objection and protest has been made by the people. But let’s first understand what BT-Brinjal is?  BT Brinjal stands for Bacillus Thuringiensis Brinjal. To understand what is BT Brinjal, first you have to understand what kind of Brinjal in India and there production. There are as many as 2000 varieties are available in India. But most of the crops are destroyed due to insect, resulting in huge loss for farmers. Bt Brinjal is a transgenic brinjal created by inserting a gene (Cry 1Ac) from the soil bacterium Bacillus thuringiensis (Bt) into Brinjal. This is said to give the Brinjal plant resistance against lepidopteron insects like the Brinjal Fruit and Shoot Borer and Fruit Borer. The insect's digestive processes are disrupted, ultimately resulting in its death, upon ingestion of the Bt toxin. Bacillus Thuringiensis or Bt is a soil-dwelling bacterium, which acts as a pesticide and increases brinjal yields. These seeds are also made in India. In one of the famous seed company that’s Mahyco in collaboration with American multinational Monsanto. So that why they have to produce this type of seeds. = = = = After the pressure of public, government has decided to ban BT brinjal for some time like 6 months. And after all investigation and research government will decide to remove or continue the ban. If it is not harmful for people, than government will remove the ban. I also think the same that government should ban BT Brinjals. There side effects are many and a need which has too much side effect I will not eat. I think government should not remove ban from it. I think that more awareness of BT Brinjal should be created. I am happy to see that my friends and brother sister are active and also protest against wrong. This shows a sign of development……. = = =05/03/10 Chemistry Experiment 5 =
 * What is BT Brinjal? **
 * How it effect on Health? **
 * 1) Bt brinjal appears to have 15 percent less calories and different alkaloid content compared to non GM brinjal. It contains 16 to 17 mg/kg Bt insecticide toxin. When fed to animals, effects were observed on blood chemistry with significant differences according to the sex of the animal or period of measurement. Other effects were on blood clotting time, total bilirubin, and alkaline phosphate in goats and rabbits.[[image:gmfoods.jpg width="278" height="272" align="right"]]
 * 2) Rats fed Bt brinjal had diarrhoea, increased water consumption; decrease in liver weight, and liver to body weight. Feed intake was modified in broiler chickens.
 * 3) Bt brinjal produces a protein in the vegetable cells that induce antibiotic resistance. This is recognized as a major health problem and is inappropriate for commercial use. Changes in lactating cows were observed in increased weight gain, intake of more dry roughage matter and milk production up by 10 to 14 percent as if they were treated by a hormone.
 * 4) Environment activists says the effect of GM (genetically modified) crops on rats have shown to be fatal for lungs and kidneys. It is dangerous to introduce these experimental foods into the market without proper research, they say.
 * 5) The side effect of BT cotton can also be seen now days.
 * BT Brinjal Controversy and my **[[image:against-bt-brinjal.jpg width="253" height="239" align="right"]]** views **

** (Amount Of Heat Given In A Exothermic Redox Reaction) **
** Aim: **  To find the amount of heat produce in a exothermic, displacement, redox reaction between CuSO4 and Zn and also its uncertainity. ** Things Required: **  Beam Balance, Beaker, Dropper, thermometer and measuring cylinder, stop watch, Motar and pestle. 
 * Reagent Required:** CuSO4 (1 MOL) and Zn 
 * Procedure:**
 * Take 4 g of Zn and crush them with the help of motar and pestle
 * Than put 20 ml of CuSO4 in a beaker with help of measuring cylinder.
 * measure the intial temperature of CuSO4 the reagents.
 * Put themometer in a beaker and a stop watch near to it.
 * Put CuSO4 and Zn in that beaker only, as soon as you put the reagent start the stop watch.
 * Than measure the temperature at every 30s interval.
 * than find out the amount of heat given by formula heat = mass of reagent * specific heat * defference in temperature / number of moles.


 *  Observation: **heat = mass of reagent * specific heat * defference in temperature / number of moles.

Mass 100 ml = 100 g so, 20 ml = 20 g Final mass = 20 g (we are taking mass of only CuSO4 because in the formula we take the mass of liquid which is absorbing or releasing heat)

intial temp of CuSO4 = 24 C max temp by CuSO4 = 45 C difference in temp = 21C**
 * Temperature

number of moles = molarity * volume
 * Number of moles **

there fore volume = 20 cm3 means 20/1000 dm3 <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;"> molarity or ratio of all the chemicals is same so we take moles of any one reagent. <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">number of moles = molarity * volume number of moles = 1 * 20/1000 number of moles =0.02 mol

specific heat (given) = 4.18 J/(g C) <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">difference in mass or mass of chemicals = 20 g **difference in temp. = 21 C** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">number of moles =0.02 mol heat = mass of reagent * specific heat * defference in temperature / number of moles. heat = 20 * 4.18 * 21 / 0.02 heat = 87.78 KJ
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 132%; text-align: left;">Heat Given Out **

<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Uncertaniity of heat given out = unctertainity % of mass + uncertanity % of temperature + unceratnity % of number of moles
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 132%; text-align: left;">Uncertainity **

we are not taking uncreatanity % of specific heat because it is already given, so its accurate. <span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 120%; text-align: left;">**Unctertainity % of mass** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">unctertainity % of mass= (error/ actual value)*100 (error is the least count) unctertainity % of mass= (0.5/ 20 )*100 unctertainity % of mass= 2.5 % <span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 120%; text-align: left;">**Unctertainity % of temperature** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">uncertanity % of temperature= (error/ actual value)*100 uncertanity % of temperature= (1/21)*100 uncertanity % of temperature= 4.76% <span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 120%; text-align: left;">**Unctertainity % of number of moles** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">uncertanity % of number of moles = uncertanity % of molarity + uncertanity % of volume molarity was given to us, so we have to find the uncertanity of volume

uncertanity % of volume= (error/ actual value)*100 uncertanity % of volume= (0.5/20)*100 uncertanity % of volume= 2.5%

Uncertaniity of heat given out = 2.5% + 4.76% + 2.5% Uncertaniity of heat given out = 9.76% So uncertanity in haet given out = 87.78 KJ * 9.76% uncertanity in haet given out = 87.78 KJ + - 8.6 KJ Precaution: **
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Total Uncertanity **<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Uncertaniity of heat given out = unctertainity % of mass + uncertanity % of temperature + unceratnity % of number of moles
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Measure the readings and least count carefully without any error.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Handle the chemicals carefully.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Wear gloves and lab court.

=28/02/10 Chemistry Experiment 4 =
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Conclusion: ** heat given out = 87.78 KJ + - 8.6 KJ

** (Amount Of Heat Given In A Exothermic Neutralization Reaction ) **
<span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Aim: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> To find the amount of heat produce in a exothermic neutralization reaction between HCl and NaOH and also its uncertainity. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Things Required: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> Beam Balance, Beaker, Dropper thermometer and measuring cylinder. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">
 * Reagent Required:** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">HCl (2 MOL) and NaOH (2 MOL) <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">
 * Procedure:**
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">measure the initial mass of beaker (A) without anything with the help of beam balance. [[image:6C388-exothermic-reaction.gif width="119" height="188" align="right"]]
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Than put 25 cm3 of HCl in another beaker with help of measuring cylinder.
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Than put 25 cm3 of NaOH in another beaker with help of measuring cylinder.
 * measure the intial temperature of both the reagents.
 * in beaker (A) put a thermometer and add both the reagent in a beaker.
 * measure the final temperature of the netralization reaction, till it ends.
 * measure the final mass of beaker (A) without thermometer.,
 * than find out the amount of heat given by formula heat = mass of reagent * specific heat * defference in temperature / number of moles.

<span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Mass <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 121%; text-align: left;">initial mass without anything = 33.06 g final mass = 82.94 g <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">difference in mass or mass of chemicals = 49.88 g intial temp of HCl = 25 C intial temp of NaOH = 25 C final temp when both were mixed = 35 C difference in temp. = 10 C**
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;"> Observation: **
 * Temperature

<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">**number of moles = molarity * volume**
 * Number of moles **

there fore volume = 25 cm3 means 25/1000 dm3 <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;"> molarity of all the chemicals is same so we take moles of any one reagent. <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">number of moles = molarity * volume number of moles = 2 * 25/1000 number of moles =0.05 mol

specific heat (given) = 4.18 J/(g C) <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">difference in mass or mass of chemicals = 49.88 g difference in temp. = 10 C <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">number of moles =0.05 mol heat = mass of reagent * specific heat * defference in temperature / number of moles. heat = 49.88 * 4.18 * 10 / 0.05 heat = 41.7 KJ
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 132%; text-align: left;">Heat Given Out **

<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Uncertaniity of heat given out = unctertainity % of mass + uncertanity % of temperature + unceratnity % of number of moles
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 132%; text-align: left;">Uncertainity **

we are not taking uncreatanity % of specific heat because it is already given, so its accurate. <span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 120%; text-align: left;">**Unctertainity % of mass** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">unctertainity % of mass= (error/ actual value)*100 (error is the least count) unctertainity % of mass= ( 0.02/49.88)*100 unctertainity % of mass= 0.04% <span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 120%; text-align: left;">**Unctertainity % of temperature** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">uncertanity % of temperature= (error/ actual value)*100 uncertanity % of temperature= (1/10)*100 uncertanity % of temperature= 10% <span style="color: #0000ff; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 120%; text-align: left;">**Unctertainity % of number of moles** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">uncertanity % of number of moles = uncertanity % of molarity + uncertanity % of volume molarity was given to us, so we have to find the uncertanity of volume

uncertanity % of volume= (error/ actual value)*100 uncertanity % of volume= (0.5/25)*100 uncertanity % of volume= 2%

Uncertaniity of heat given out = 0.04% + 10% + 2% Uncertaniity of heat given out = 12.04% So uncertanity in haet given out = 41.7 KJ * 12.04% uncertanity in haet given out = 41.7 KJ + - 5.02 KJ Precaution: **
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Total Uncertanity **<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Uncertaniity of heat given out = unctertainity % of mass + uncertanity % of temperature + unceratnity % of number of moles
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Measure the readings and least count carefully without any error.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Handle the chemicals carefully.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Wear gloves and lab court.


 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Conclusion: ** heat given out = 41.7 KJ + - 5.02 KJ

=18/02/10 Chemistry Experiment 3 =

** (Reaction Between Calcium Carbonate and Hydro Chloric Acid) **
<span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Aim: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> To find the difference in time to reach the emission of carbon dioxide gase till 100000 pps with different masses of calcium carbonate. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Things Required: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> Beam Balance, Data Logos with carbon dioxide sensor, Laptop, Beaker, Spatula and measuring cylinder. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">**Reagent Required:** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Calcium carbonate and HCl Acid (1 mol) <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Prediction: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> The more we put calcium carbonate in constant HCl it take more less time to emit 100000 pps carbon dioxide. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">**Procedure:** Observation: **<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Taking masses of calcium carbonate 0.5 g 1 g 1.5 g2 g 2.5 g and 50 cm3 of HCl <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Install Data logos software in your laptop.
 * measure the different masses of calcium carbonate and record it, atleast take 3
 * Then take 50 cm3 of HCl and in a beaker. <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">[[image:chfa_01_img0184.jpg width="213" height="315" align="right"]]
 * Prepare the data logos with a carbon dioxide sensor and connect it to laptop.
 * Put the HCl in CO2 sensor beaker.
 * Put calcium carbonate in that beaker.
 * Than immediately put CO2 sensor in that beaker and immediately start collect in your laptop in data logos software.
 * Graph will be plotted on your laptop.
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Then use different mass of calcium carbonate with 50 cm3 of HCl, so different graph will be plotted, repeat this at least 3 times with different masses.
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">

**<span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">

Precaution: **<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Handle lab pro carefully.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">Measure the mass of calcium carbonate.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">The black part of carbon dioxide sensor should not touch the chemicals in the beaker.
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Clean carbon dioxide sensor with tissue paper after the experiment.


 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Conclusion: **<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> The experiment worked. The more we use calcium carbonate, the more less time it take to emit 100000 pps of CO2.


 * 02/02/10**

=**<span style="color: #0000ff; font-family: Verdana,Geneva,sans-serif; font-size: 150%;">Indore and near by area is in danger **= Indore is great city every thing is possible in Indore. Weather its food or anything. Now days Indore is at the peak of development. This development can or could be stop by other near areas of Indore like Pithampur. But, how it’s possible? This could be possible because in Pithampur the poisonous waste of factories would be burned in Pithampur due to which Indore would be in danger. Because this waste has taken lot life of people in Bhopal and I think so it will take in Indore too. There is very big tragedy happened in Bhopal due this waste. Let’s first know what that tragedy was. Introduction Around 1 a.m. on Monday, the 3rd of December, 1984, in a densely populated region in the city of Bhopal, Central India, a poisonous vapor burst from the tall stacks of the Union Carbide pesticide plant. This vapor was a highly toxic cloud of methyl isocyanate. Of the 800,000 people living in Bhopal at the time, 2,000 died immediately, and as many as 300,000 were injured. In addition, about 7,000 animals were injured, of which about one thousand were killed. “A series of studies made five years later showed that many of the survivors were still suffering from one or several of the following ailments: partial or complete blindness, gastrointestinal disorders, impaired immune systems, post traumatic stress disorders, and menstrual problems in women. This incident we now refer to as the Bhopal Gas Tragedy, which has also been called “Hiroshima of the Chemical Industry” one of the worst commercial industrial disasters in history. The Tragedy: Possible Causes The post-accident analysis of the process showed that the accident started when a tank containing methyl isocyanate (MIC) leaked. MIC is an extremely reactive chemical and is used in production of the insecticide carbaryl. It is presumed that the scientific reason for the accident at Bhopal is that water entered the tank where about 40 cubic meters of MIC was stored. When water and MIC mixed, an exothermic chemical reaction started, producing a lot of heat. As a result, the safety valve of the tank burst because of the increase in pressure. This burst was so violent that the coating of concrete around the tank also broke. It is presumed that between 20 and 30 tonnes of MIC were released during the hour that the leak took place. The gas leaked from a 30 m high chimney and this height was not enough to reduce the effects of the discharge. The reason was that the high moisture content (aerosol) in the discharge when evaporating, gave rise to a heavy gas which rapidly sank to the ground. The weather egged on this process. The conditions on the fateful day were typical for a clear night in the region, with a weak wind which frequently changed direction, which in turn helped the gas to cover more area in a shorter period of time (about one hour). The weak wind and the weak vertical turbulence caused a slow dilution of gas and thus allowed the poisonous gas to spread over considerable distances.

Why Pithampur is in trouble?
It’s in danger because the remaining waste around 40 tones of it would be burn in Pithampur in a incinerator which would be install by government. The part of this waste of try to dump in a land but it was not decomposing and was harming the environment and the land. This dumped waste has also poisoned the water near by it. Than government decide to send this waste in Gujarat and burn the waste there in incinerator but than Gujarat government refused to burn the waste. Than M.P. government decide to burn the waste in Pithampur. Because our state was having an incinerator that’s why it was decided that the waste would be send to Gujarat but as Gujarat government refused than our government decided to install an incinerator in Pithampur. The waste should be burned properly in incinerator, if not again a big tragedy would happen. The waste transportation should be also properly.

Pick of an inconerator

God’s now what would happen to my land (Indore) and about me too. = = =24/01/10 Chemistry Experiment 2 =

** (Endothermic Reaction between Sodium Bicarbonate and Citric Acid) **
<span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Aim: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> To find the difference in temperature, when different molarity of Citric acid was added to Sodium Bicarbonate. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Things Required: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> Beam Balance, Data Logos with temperature sensor, Laptop, Beaker, Spatula and Water. <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">**Reagent Required:** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Sodium Bicarbonate and Citric Acid <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">** Prediction: ** <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> As it’s endothermic reaction so, it absorb the heat, therefore the more we use the molarity of Citric acid, the more the temperature goes down.

<span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">**Procedure:** Observation:
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Install Data logos software in your laptop.
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Prepare different molarity of citric acid using different mass of it and 100 cm3 of water. At least take 5 different molarity.
 * Then take 60 cm3 of a molarity in beaker.
 * Prepare the data logos with a temperature sensor and connect it to laptop.
 * Put the temperature sensor in beaker.
 * Then in your laptop, in data logos software start collect.
 * After 2 or 3 sec put 5 gm of sodium bicarbonate in a beaker.
 * Graph will be plotted on your laptop.
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Then use different molarity different graph will be plotted, repeat this at least 5 times with different molarity
 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">

Precaution: **
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Make molarity carefully.
 * <span style="color: #000000; font-family: Arial,Helvetica,sans-serif;">The black part of temperature sensor should not touch the chemicals in the beaker.
 * <span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;">Clean temperature sensor with tissue paper after the experiment


 * <span style="color: #ff0000; display: block; font-family: Arial,Helvetica,sans-serif; font-size: 110%; text-align: left;">Conclusion: **<span style="color: #000000; display: block; font-family: Arial,Helvetica,sans-serif; text-align: left;"> The experiment worked. The more we use the molarity of Citric acid, the more the temperature goes down.

=22/01/10 1) Experiment With Salt And Water =

**Activity:** Amount of salt disolve in water at different temperature.

**Things required:** 4 Beakers, Salt (NaCl), water, top balance

The more the temperature is high the more the salt disolves because rate of reaction depend upon temperature.
 * Prediction **:

**Procedure:**
 * Take atleast 4 beakers and measure its mass with the help of top balance.
 * Than put 100 cm3 of water in each.
 * Than heat the water at different temperature
 * After it start adding the salt in little quantity, till it disolve.
 * Than measure the mass of each beaker.
 * The difference in mass will the mass of the salt disolved in the water.



(30*C) || Experiment 3 (40*C) ||
 * Observation** :
 * Temperature || Experiment 1 (19*C) || Experiment 2
 * Initial mass of beaker (gm) || 152.27 || 152.08 || 125.46 ||
 * Final mass of beaker (gm) || 157.40 || 164.39 || 142.12 ||
 * Difference or mass of salt dissolved (gm) || 5.13 || 12.31 || 16.66 ||


 * Precaution: **
 * Heat the water carefully.
 * As soon you heat the water till particular temperature, than start adding the salt suddenly.
 * Put the salt in very small quantity.

**Conclusion:** The more the water is heated the more the salt is disolved. Hence the prediction was right.