Benzene burns in air to produce carbon dioxide and water...?
a million. the chemical formula for Benzene is C6H6.. which assertion approximately benzene is acceptable? a. its a hydrocarbon (fragrant hydrocarbon) 2. The equation under isn't balanced.. 2C6H6 + ........O2 --> 12CO2 + 6H2O what number oxygen molecules could be required to stability the equation? b. 15 (24 + 6 = 30 oxygen atoms or 15 O2 molecules) 3. Benzene burns with an extremely smoky flame, producing a large number of soot. it fairly is via the fact.... c. it has a intense ratio of carbon to hydrogen atoms 4. the fee of benzene is increasing. this could be by using fact. d. the expenses of crude oil and coal are increasing 5. what's an benefit of ethanol as a gasoline while it burns? b. carbon dioxide and water are produced. c. no sulfur dioxide is produced. i think of that there are 2 reward to employing ethanol. the 1st, of direction, is that no sulfur dioxide is produced. Sulfur dioxide provides upward push to acid rain and is troublesome for asthmatics. Secondly, it burns producing basically carbon dioxide and water that's named sparkling burning.
Ethanol, C2H5OH, burns with the oxygen in air to give carbon dioxide and water. (rest in bottom)?
This Site Might Help You. RE: Ethanol, C2H5OH, burns with the oxygen in air to give carbon dioxide and water. (rest in bottom)? Write a balanced equation for this reaction. How many moles of water are produced when 0.69 mol C2H5OH are used?
Ethanol undergoes combustion in oxygen to produce carbon dioxide gas and liquid water.?
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How do you calculate the mass of carbon atoms in CO2?
This question is meaningless. It is meaningful to talk about the mass of standalone atoms of a particular isotope of carbon, and it is meaningful to talk about the mass of individual molecules of CO₂, which is NOT the same as the sum of the mass of one carbon atom and twice the mass of two oxygen atoms, because the chemical bonding of the three atoms together to form the molecule gives off what it called binding energy. Because of the Einsteinian relationship of mass and energy, that loss of energy from the molecule corresponds to a loss of mass. You cannot validly attribute that loss of mass to any one component atom, nor can you validly split somehow the loss of mass for the molecule among the various component atoms..
Chemistry - Specific Heat, Molar Heat Combustion?
Please show me how to work out these 3 problems, i'm stuck! Thank you. 1. A quantity of 1.834 g of glycine (C2H5NO2) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose by 2.29C. If the heat capacity of the bomb plus water was 10.4 kJ/°C, calculate the molar heat of combustion of glycine (C2H5NO2). ______ kJ/mol 2.A 34.82 g stainless steel ball bearing at 117.82°C is placed in a constant-pressure calorimeter containing 120.0 mL of water at 18.44°C. If the specific heat of the ball bearing is 0.474 J/g·°C, calculate the final temperature of the water. Assume the calorimeter to have negligible heat capacity. ____ Celsius 3.Ethanol (C2H5OH) burns in air to produce carbon dioxide and liquid water. Calculate the heat released (in kilojoules) per gram of the compound reacted with oxygen. The standard enthalpy of formation of ethanol is -277.7 kJ/mol. ________ kJ/g C2H5OH
Calculate the heat released (in kilojoules) per gram of the compound Octane reacted with oxygen.?
You need to look up additional data to do this problem, so I'll show you how to do it and you can do the details: First, write the balanced chemical reaction: C8H18 + 25/2 O2 -> 8 CO2 + 9 H2O Find the heat of reaction (enthalpy of reaction): Here is where you need to look up additional data. The heat of reaction is given by: Enthalpy of reaction = Sum of [(enthalpies of formation of products)*(coefficient in balanced chemical reaction)] - Sum of [(enthalpies of formation of reactants)*(coefficient in balanced chemical reaction)] You need to look up the enthalpies of formation of the other chemicals in the reaction. The result of that calculation will give you the enthalpy of reaction per mole of octane. Then you figure out how many moles of octane are in 1 gram. Finally multiply that value by the enthalpy of reaction per mole. Hope this helps.
What is the balanced chemical equation for the combustion of propane?
Start with the equation. 1 C3H8 + 5 O2 --> 3 CO2 + 4 H2O rewrite CCCHHHHHHHH + OO --> COO + HHObalance carbons OO COO HHObalance hydrogens OO COO HHObalance oxygens OO HHO double check LH and RH side OO add columns __1__ __5__ __3__ __4__ return to the original equation and fill in the coefficients. By keeping the atoms of the molecules connected to form molecules, most chemistry students make fewer errors. If the equation doesn't balance the first time simply going right to left try again and look at the molecules that contain the same elements on the same side of the equation arrow. (CO2 and H2) in this case. Sorry for not being able to format this better.
What is the chemical equation for burning sugar?
I presume you are looking for the equation for the combustion of sugar? The general formula for sugar is given as [math]C_{11}H_{22}O_{11}[/math]. Assuming complete combustion, you would end up with:[math]C_{12}H_{22}O_{11}+ 12O_2 \rightarrow 12CO_2+ 11H_2O [/math]Of course, complete combustion does not occur with sugar. Mix this with potassium nitrate and you have a vigorous home-made smoke bomb (this is a nice touch to add to that home-made volcano science project). After burning, you will be left with a black mass. This also happens if you just try to set sugar on fire. The actual chain of reactions is quite complex and what you actually end up with is a whole bunch of aldehydes, ketones, carboxylic acids and more forming.That being said, this is actually a pretty dangerous reaction, making sugar one of the more dangerous items sitting in your pantry. If you increase the surface area of the sugar by milling it very fine, you can further accelerate this reaction to the point where it is fatal. In February 2008, dust clouds of fine sugar lead to explosions killing 14 people at the Imperial Sugar manufacturing facility in Port Wentworth, Georgia.It kills you fast and it kills you slow (diabetes!). Sugar may be the most deadly food known to man.
Combustion of ethanol and CO2 Question?
Step 1: Write equation and balance C2H5OH (l) + O2(g) --------> CO2(g) + H2O(l) C2H5OH(l) + 3O2(g) ---------> 2CO2(g) + 3H2O(l) Step 2: Calculate the number of grams of ethanol using the Density formula: Mass Ethanol = Density x volume = 0.789g/ml x 4.63ml = 3.65g Step 3: Calculate Molar Mass for ethanol and oxygen C2H5OH = 2(12.01) + 6(1.008) + 16 = 46.07g O2 = 2(16) =32g H2O = 2(1.008) + 16 = 18.02g Note: Oxygen is a diatomic element Step 4: Find the limiting reactant. 3.65g C2H5OH (1 mol C2H5OH / 46.07g C2H5OH) = .0792 mol C2H5OH 15.50g O2 (1 mol O2 / 32g O2) = .4844 mol of O2 Since, number of Ethanol moles is less than the Oxygen moles, ethanol is the limiting reactant. Step 5. Calculate the amount of water produced (Theoretical Yield). .0792 mol C2H5OH (18.02g H20 / 1 mol C2H5OH)(3 mol H2O / 1 mol C2H5OH) = 4.28g H2O Step 6. Find the amount of water collected in grams using the density formula. Mass H20 = Density x Volume = 1.00g/ml x 3.70ml = 3.70g H2O Step 7. Calculate the percent yield of H2O. %Yield = (Actual Yield/Theoretical Yield) x 100% = 3.70g/4.28g x 100 = 86.4% Bill
What is the balanced chemical equation for the combustion of propane?
Start with the equation. 1 C3H8 + 5 O2 --> 3 CO2 + 4 H2O rewrite CCCHHHHHHHH + OO --> COO + HHObalance carbons OO COO HHObalance hydrogens OO COO HHObalance oxygens OO HHO double check LH and RH side OO add columns __1__ __5__ __3__ __4__ return to the original equation and fill in the coefficients. By keeping the atoms of the molecules connected to form molecules, most chemistry students make fewer errors. If the equation doesn't balance the first time simply going right to left try again and look at the molecules that contain the same elements on the same side of the equation arrow. (CO2 and H2) in this case. Sorry for not being able to format this better.