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77julia77 [94]
3 years ago
10

A balloon has a volume of 1.40 L at 24.0ºC. The balloon is heated to 48.0°C. Calculate the new volume of the balloon.

Chemistry
1 answer:
ruslelena [56]3 years ago
5 0

Answer:I am not sure how

Explanation: sorry free trial

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When 8.0 g H₂ react with 8.0 g O₂ in the reaction 2H₂ + O₂ → 2H₂O, what are the theoretical yield and the limiting reactant?
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Now, we have to determine the limiting reagent.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g of

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂O

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.

Now, we have to determine the limiting reagent.4 g of H₂ reacts with 32 g of O₂ 1 g of H₂ reacts with 32/4 g of O₂ 3 g of H₂ reacts with 32/4 x 3 = 24 g ofBut according to the question, 29 g of O₂ is present. 2So, the limiting reactant is hydrogen.Now, 4 g of H₂ forms 36 g of H₂O1 g of H₂ forms 36/4 g of H₂O. 3 g of H₂ forms 36/4 x 3 = 27 g of H₂OMaximum amount of water that can be formed is 27 g.For, amount of oxygen left of unreacted, Only 24 g of oxygen will react.But 29 g is the given amount. Amount of oxygen unreacted = 29 - 24 = 5 g

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