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4 years ago

Liquid butane, C4H10, is stored in cylinders to be used

Chemistry

1 answer:

Ganezh [65]4 years ago

3 0

Answer:

13.0kJ

Explanation:

The heat to vaporize a liquid is equal to the amount of liquid in moles multiplied by the specific heat of vaporiztion per mole.

First, calculate the number of moles in 35.5g of butane.

Molar mass of butane: 58.124 g/mol

Number of moles = mass in grams/molar mass

Number of moles = 35.5g / 58.124g/mol = 0.6107632mol

Now, calculate the heat to vaporize that amount of liquid butane:

Heat = number of moles × specific heat of vaporization

Heat = 0.6107632mol × 21.3kJ/mol = 13.0 kJ

The answer must be reported with 3 significant figures.

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A solution is made by dissolving 4.35 g of glucose (C6H1206) in 25.0 mL of water at 25 °C.Calculate the molality of glucose in t

Roman55 [17]

Answer:

The molality is unchanged (0.96 molal)

Explanation:

Step 1: Data given

mass of glucose = 4.35 grams

volume of water = 25.0 mL

Density of water = 1.00 g/mL

Molar mass of glucose = 180.156 g/mol

Step 2: Calculate number of moles

moles of glucose = mass of glucose / Molar mass of glucose

moles of glucose = 4.35 grams / 180.156 g/mol

moles of glucose = 0.024 moles

Step 3: Calculate mass of water

mass = density * volume

mass of water = 1.00 g/mL * 25.0 mL

mass of water = 25 g = 0.025 kg

Step 4: Calculate molality

molality = Number of moles / mass of water

molality = 0.024 moles / 0.025 kg

molality = 0.96 molal

Suppose you take a solution and add more solvent, so that the original mass of solvent is doubled.

This means double mass of water = 2*0.025 kg = 0.050 kg

Now molality is 0.024 moles / 0.050 kg = 0.48 molal

When the mass of solvent is doubled, the molality is halved from 0.96 molal to 0.48 molal

You take this newsolution and add more solute, so that the original mass of the solute is doubled.

This means double mass of glucose = 2*4.35 g = 8.70 g

8.70 grams of glucose = 8.7 grams * 180.156 g/mol = 0.048 moles

molality = 0.048 moles / 0.050 kg = 0.96 molal

The molality is unchanged

5 0

4 years ago

Give examples to three solutions you commonly use and identify the solvent and solutes that make it each up.

8_murik_8 [283]

Mouthwash:
solvent - water
solute - alcohols

vinegar:
solvent - water
solute - acetic acid

bleach:
solvent - water
solute - sodium hypochlorite

hope this helps!!

8 0

3 years ago

Read 2 more answers

The equilibrium constant Kp for the reaction I2(g) + Br2(g) ⇀↽ 2 IBr(g) + 11.7 kJ is 280 at 150◦C. Suppose that a quantity of IB

mixas84 [53]

Answer:

$\large \boxed{\text{0.0120 atm }}$

Explanation:

The balanced equation is

I₂(g) + Br₂(g) ⇌ 2IBr(g)

Data:

Kp = 280

p(IBr) = 0.200 atm

1. Set up an ICE table.

Let p = the initial pressure of IBr. Then

$\begin{array}{ccccccc}\rm \text{I}_{2}& + & \text{Br}_{2} & \, \rightleftharpoons \, & \text{2IBr} & & \\0 & & 0 & & p & & \\+x & & +x & & -2x & &\\x & & x} & & 280 & & \\\end{array}$

2. Calculate p(I₂)

$\begin{array}{rcl}K_{\text{p}}&=&\dfrac{p_{\text{IBr}}^{2}} {p_{\text{I}_2}^{2}}\\\\280&=&{\dfrac{0.200^{2}}{x^{2}}&&\\\\280x^{2} & = &0.0400\\x^{2} & = &\dfrac{0.0400}{280 }\\\\& = & 1.429 \times 10^{-4}\\x & = & \textbf{0.0120 atm}\\\end{array}\\\text{The partial pressure of iodine is $\large \boxed{\textbf{0.0120 atm }}$}$ }