Express your answer using two significant figures. 1.7 km^2

Butane, C4H10 burns in oxygen. How many grams of water vapor, H2O, are produced by the combustion of 580 grams of butane at stan

zepelin [54]

Answer:

The answer to your question is 900 g of water vapor

Explanation:

Data

mass of H₂O = ?

mass of butane = 580 g

Balanced chemical reaction

2C₄H₁₀ + 13O₂ ⇒ 8CO₂ + 10H₂O

Process

1.- Calculate the molar weight of butane and water

Butane (C₄H₁₀) = 2[(12 x 4) + (1 x 10)]

= 2[48 + 10]

= 2[58]

= 116 g

Water (H₂O) = 10[(1 x 2) + (1 x 16)]

= 10[2 + 16]

= 10[18]

= 180 g

2.- Use proportions and cross multiplication to find the mass of water vapor

116 g of butane ------------- 180 g of water

580 g of butane ---------- x

x = (580 x 180) / 116

x = 900 g of water vapor

3 0

3 years ago

What is the total probability of finding a particle in a one-dimensional box in level n = 4 between x = 0 and x = L/8?

Lubov Fominskaja [6]

Answer:

P = 1/8

Explanation:

The wave function of a particle in a one-dimensional box is given by:

$\psi = \sqrt \frac{2}{L} sin(\frac{n \pi x}{L})$

Hence, the probability of finding the particle in the one-dimensional box is:

$P = \int_{x_{1}}^{x_{2}} \psi^{2} dx$

$P = \int_{x_{1}}^{x_{2}} (\sqrt \frac{2}{L} sin(\frac{n \pi x}{L}))^{2} dx$

$P = \frac{2}{L} \int_{x_{1}}^{x_{2}} (sin^{2}(\frac{n \pi x}{L}) dx$

Evaluating the above integral from x₁ = 0 to x₂ = L/8 and solving it, we have:

$P = \frac{2}{L} [\frac{L}{16} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})]$

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{n \pi}{4})}{n \pi})$

Solving for n=4:

$P = \frac{1}{8} (1 - 4\frac{sin(\frac{4 \pi}{4})}{4 \pi})$

$P = \frac{1}{8} (1 - \frac{sin (\pi)}{\pi})$

$P = \frac{1}{8}$

I hope it helps you!

7 0

3 years ago

A 544 mg of a mixture of fluorene and benzoic acid was weighed out and subjected to an extraction and recrystallization. After t

butalik [34]

Answer: The percent composition of fluorene is 33.08 % and benzoic acid is 36.03 %.

Explanation:

We are given:

Mass of mixture = 544 mg

Mass of fluorene after purification = $1.80\times 10^2mg=180mg$

Mass of benzoic acid after purification = 196 mg

To calculate the percentage composition of a substance in a mixture, we use the equation:

$\%\text{ composition of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

For fluorene:

$\%\text{ composition of fluorene}=\frac{180}{544}\times 100=33.08\%$

Hence, the percent composition of fluorene is 33.08 %

For benzoic acid:

$\%\text{ composition of benzoic acid}=\frac{196}{544}\times 100=36.03\%$

Hence, the percent composition of benzoic acid is 36.03 %

5 0

4 years ago

Push your palms against a wall. What happened?

Jlenok [28]

You push yourself away

7 0

3 years ago

Silver nitrate can be produced by dissolving metallic silver in concentrated nitric acid, with an addition product being hydroge

geniusboy [140]

Answer:

2Ag(s) + 2HNO₃ (aq) → 2AgNO₃(aq) + H₂(g)

Explanation:

Metallic silver = Ag (s)

Concentrated nitric acid = HNO₃ (aq)

Product being hydrogen gas = H₂(g)

2Ag(s) + 2HNO₃ (aq) → 2AgNO₃(aq) + H₂(g)

Be careful AgNO₃ doesn't precipitate

4 0

3 years ago