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3 years ago

Which of the following is an example of a decomposition reaction

1 answer:

8 0

HgO--> Hg + O2 Decompostion can spilt one compound into two parts or more pieces. These pieces can be elements or simpler compounds.

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If 5 L of butane is reacted what volume of carbon dioxide is produced ILL GIVE BRAINLIEST

Len [333]

Answer: First, here is the balanced reaction: 2C4H10 + 13O2 ===> 8CO2 + 10H2O.

This says for every mole of butane burned 4 moles of CO2 are produced, in other words a 2:1 ratio.

Next, let's determine how many moles of butane are burned. This is obtained by

5.50 g / 58.1 g/mole = 0.0947 moles butane. As CO2 is produced in a 2:1 ratio, the # moles of CO2 produced is 2 x 0.0947 = 0.1894 moles CO2.

Now we need to figure out the volume. This depends on the temperature and pressure of the CO2 which is not given, so we will assume standard conditions: 273 K and 1 atmosphere.

We now use the ideal gas law PV = nRT, or V =nRT/P, where n is the # of moles of CO2, T the absolute temperature, R the gas constant (0.082 L-atm/mole degree), and P the pressure in atmospheres ( 1 atm).

V = 0.1894 x 0.082 x 273.0 / 1 = 4.24 Liters.

Explanation:

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3 years ago

What does 3.98 g/cm”3 equal

Elenna [48]

Answer:

ρ = 1000 kg/m3

Explanation:

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4 years ago

How many grams are equivalent to 20 moles of water?

Eduardwww [97]

Answer: 360.3056 grams i think

Explanation:

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3 years ago

2. Which of the following is TRUE about the insect spray? *

natali 33 [55]

B is the answer. It just depends on the area you are in.

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3 years ago

Read 2 more answers

Freeze-drying is a process used to preserve food. If strawberries are to be freeze-dried, then they would be frozen to -80.00 °C

katrin2010 [14]

Answer:

1. 389 kJ; 2. 7.5 µg; 3. 6.25 days

Explanation:

1. Energy required

The water is converted directly from a solid to a gas (sublimation).

They don't give us the enthalpy of sublimation, but

$\Delta_{\text{sub}}H = \Delta_{\text{fus}}H + \Delta_{\text{vap}}H = 6.01 + 40.68 = 46.69 \text{ kJ}\cdot\text{mol}^{-1}$

The equation for the process is then

Mᵣ: 18.02

46.69 kJ + H₂O(s) ⟶ H₂O(g)

m/g: 150

(a) Moles of water

$\text{Moles} = \text{150 g} \times \dfrac{\text{1 mol}}{\text{18.02 g}} = \text{8.324 mol}$

(b) Heat removed

46.69 kJ will remove 1 mol of ice.

$\text{Heat removed} = \text{8.234 mol} \times \dfrac{\text{46.69 kJ}}{\text{1 mol}} = \textbf{389 kJ}\\\text{It takes $\large \boxed{\textbf{389 kJ}}$ to remove 150 g of ice}$

2. Mass of water vapour in the freezer

For this calculation, we can use the Ideal Gas Law — pV = nRT

(a) Moles of water

Data:

$p = 1.00 \times 10^{-3}\text{ torr } \times \dfrac{\text{1 atm}}{\text{760 torr}} = 1.316 \times 10^{-6}\text{ atm}$

V = 5 L

T = (-80 + 273.15) K = 193.15 K

Calculation:

$\begin{array}{rcl}pV & = & nRT\\1.316 \times 10^{-6}\text{ atm} \times \text{5 L} & = & n \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times \text{193.15 K }\\6.6 \times 10^{-6} & = & 15.85n\text{ mol}^{-1} \\n & = & \dfrac{6.6 \times 10^{-6}}{15.85\text{ mol}^{-1}}\\\\& = & 4.2 \times 10^{-7} \text{ mol}\\\end{array}$

(b) Mass of water

$\text{Mass} = 4.2 \times 10^{-7} \text{ mol} \times \dfrac{\text{18.02 g}}{\text{1 mol}} = 7.5 \times 10^{-6}\text{ g} = 7.5 \, \mu \text{g}\\\\\text{At any given time, there are $\large \boxed{\textbf{7.5 $\mu$g}}$ of water vapour in the freezer.}$

3. Time for removal

You must remove 150 mL of water.

It takes 1 h to remove 1 mL of water.

$\text{Time} = \text{150 mL} \times \dfrac{\text{1 h}}{\text{1 mL}} = \text{150 h} = \text{6.25 days}$

5 0

4 years ago