The answer to this question will depend on what the substance that is reacting is. You will need to find the appropriate standard enthalpy value, which corresponds to the amount of enthalpy change when one mole of matter is transformed by a chemical reaction in standard conditions.
q = mHvq = heatm = mass (30g)Hv = heat of vaporization (2,260J/g)q = (30g)(2,260J/g)q = 67,800JWhen 30 grams of water is converted into steam, how much heat is absorbed?67,800J of heat, also represented as 67.8kJ of heat is absorbed.
46.2 kJ/mole of heat is a part of the synthesis of 1.0 mole of ammonia. As a result, 231 kJ/mole of heat will result in the synthesis of 5.0 moles of ammonia.
Approx. 15 kJ.
I assume by " much " you mean grams tin. Balanced equation first.Sn + 4HCl -> SnCl4 + 2H2Now, find moles HCl by......Molarity = moles of solute/Liters of solution ( 32.0 ml = 0.032 Liters )1.75 M HCl = X moles/0.032 liters= 0.056 moles HCl=============== Now,0.056 moles HCl (1 mole Sn/4 moles HCl)(118.7 grams /1 mole Sn)= 1.7 grams of tin-----------------------
The potassium hydroxide is what we call the limiting reactant. 4 moles of potassium hydroxide react with 2 moles of magnesium chloride. The third mole of magnesium chloride is in excess and has no effect.
540 calories per gram is absorbed when water vaporizes at its boiling point. Called the latent heat of vaporization. 540 x 23.1 x 18 = 224532 calories
Too much heat would result in brake damage, which is absorbed in the brake drum.
51%
Molecular weight of water is 18 so 9 moles is 162 grams. Latent heat of vaporisation is 550 cal/gram, so the amount of heat is 89100 cal or 89.1 kcal.
q = mHvq = heatm = mass (30g)Hv = heat of vaporization (2,260J/g)q = (30g)(2,260J/g)q = 67,800JWhen 30 grams of water is converted into steam, how much heat is absorbed?67,800J of heat, also represented as 67.8kJ of heat is absorbed.
No, it stays pretty much the same.
46.2 kJ/mole of heat is a part of the synthesis of 1.0 mole of ammonia. As a result, 231 kJ/mole of heat will result in the synthesis of 5.0 moles of ammonia.
23 percent of incoming solar energy is usually absorbed by the ozone. Temperature, not how much
for every 5 moles of y there is 1 mole of x, so 3.8/5 is .76
The necessary heat is 9,22 joules.
The necessary heat to transform liquid chlorine (Cl2) to a gas is 9,756 kcal.
2 kJ.