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adelina 88 [10]
2 years ago
13

calculate the volume od a CO2 cartridge that has a pressure of 850 PSI at a temperature of 21ºc the cartridge contains 0.273 mol

of CO2
Chemistry
1 answer:
gladu [14]2 years ago
3 0
Answer is: volume of CO₂ is 0,113 dm³.
Ideal gas law = pV = nRT.
p = 850 PSI = 5860543,6992 Pa.
Psi <span>is the abbreviation of pound per square inch.
T = 21</span>°C = 294,15 K.
n = 0,273 mol.
R = 8,314 J/K·mol.
V = nRT ÷ p
V = 0,273 mol · 8,314 J/K·mol · 294,15 K ÷ 5860543,6992 Pa.
V = 0,00011 m³ = 0,113 dm³.
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Glade air freshener gel “disappearing” is an example of
netineya [11]

Answer:

Vaporization

Explanation:

Vaporization is the change of a specie to the gaseous state. To 'disappear' in this case simply means to change to the gaseous state.

Substances with high vapour pressure tend to be easily converted to vapour phase. Hence if Glade air freshener gel 'disappears' easily, then it has a high vapour pressure and is easily converted to vapour (gas).

8 0
1 year ago
Substitution of an amino group on the para position of acetophenone shifts the cjo frequency from about 1685 to 1652 cm−1 , wher
cluponka [151]

Answer:

Here's what I get.

Explanation:

The frequency of a vibration depends on the strength of the bond (the force constant).

The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.

Acetophenone

Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.

p-Aminoacetophenone

The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.

The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.

p-Nitroacetophenone

The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.

As electron density moves toward C-1, the double bond character of the C=O group increases.

The bond length decreases, so the bond becomes stronger, and wavenumber  increases to 1693 cm¹.

6 0
2 years ago
We would like to make a golden standard kilogram in the shape of circular cylinder. The density of gold is 19.32 g/cm3. a) Find
iragen [17]

Explanation:

a)  Using the provided information about the density  of gold, the sample size, thickness, and the following  equations and comersion factors, find the area of   the gold leaf:

V=l \cdot w \cdot h=A \cdot h\\m=\rho \cdot V

Gold _{\rho}=19.32 \mathrm{g} / \mathrm{cm}^{3}

1 \mu=10^{-6} \mathrm{m}

First, find the volume of the sample and then find the area of the sample.

V=\frac{m}{\rho}=\frac{27.6 \mathrm{g}}{19.32 \mathrm{g} / \mathrm{cm}^{3}} \cdot\left(\frac{0.01 \mathrm{m}}{1 \mathrm{cm}}\right)^{3} =\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}

V=A \cdot h \rightarrow A=\frac{V}{h}=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{10^{-6} \mathrm{m}} \approx 1.429 \mathrm{m}^{2}

b.  Using the provided information from part a ), the radius of the cylinder, and the following equation for the volume of a cylinder, find the length of the fiber :

V=\pi r^{2} h \rightarrow h=\frac{V}{\pi r^{2}}

h=\frac{1.429 \times 10^{-6} \mathrm{m}^{3}}{\pi \cdot\left(2.5 \times 10^{-6} \mathrm{m}\right)^{2}} \approx 72778 \mathrm{m}

8 0
2 years ago
Which statement about gases is true? A. They are made up of particles that always move very slowly. B. They are made up of parti
rosijanka [135]

Answer: D. They are made up of hard spheres that are in random motion.

Explanation:

A gas is a <u>state of aggregation of matter</u> in which, under certain conditions of temperature and pressure, <u>its molecules interact weakly with each other, without forming molecular bonds</u>, adopting the shape and volume of the container that contains them and tending to separate everything possible because of its <u>high concentration of kinetic energy</u>.

The molecules of a gas are practically <u>free</u> and have the ability to be distributed throughout the space in which they are contained because <u>the gravitational forces and attraction between them are practically negligible</u> compared to the speed at which they move. .

Therefore, gas molecules do not travel specific trajectories or vibrate in a stationary position, instead <u>they move quickly and randomly through the entire space of the container that contains them.</u>

7 0
2 years ago
Read 2 more answers
If 6.00 g of the unknown compound contained 0.200 mol of C and 0.400 mol of H, how many moles of oxygen, O, were in the sample?
Arada [10]

Convert moles to mass.

mass C = 0.2 mol * 12 g / mol = 2.4 g

mass H = 0.4 mol * 1 g / mol = 0.4 g

So mass left for O = 6 g – (2.4 g + 0.4 g) = 3.2 g

 

Calculating for moles O given mass:

moles O = 3.2 g / (16 g / mol) = 0.2 moles

 

Answer:

<span>0.2 moles O</span>

8 0
2 years ago
Read 2 more answers
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