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uranmaximum [27]
2 years ago
15

A researcher suspects that the pressure gauge on a 54.3-L gas cylinder containing nitric oxide is broken. An empty gas cylinder

weighs 90.0 lb. The weight of the partially full cylinder is 116.5 lb. This cylinder is located in a relatively chilly service hallway at 287 K. Follow the steps below to use the compressibility charts to estimate the pressure of the gas and the reading that the pressure gauge should have. How many moles of nitric oxide, NO, are in the cylinder?
Chemistry
1 answer:
mixer [17]2 years ago
3 0

Answer:

Number of moles nitric acid in the cylinder is 400.539g/mol.

Explanation:

From the given,

Weight of empty gas cylinder W_{1}= 90.0 lb= 40823.3 gramsWeight of full cylinder[tex]W_{2} = 116.5 lb= 52843.511 gramsThe critical temperature = 287 KThe critical pressure 54.3 LMolar mass of nitric acid = [tex]M_{NO} = 30.01 g/mol

Number of moles nitric acid = n_{NO} =?

The mass of nitric acid in the cylinder = W_{NO}=W_{2}-W_{1}

=52843.511-40823.3 =12,020.2g

Number of moles of nitric acid =

\frac{Given\,mass}{Molar\,mass}=\frac{12,020.2}{30}=400.539g/mol

Therefore, number of moles nitric acid in the cylinder is 400.539g/mol.

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A sample of gas occupies 10 L at STP. What
puteri [66]

Pressure is 5.7 atm

<u>Explanation:</u>

P1 = Standard pressure = 1 atm

P2 = ?  

V1 = Volume = 10L

V2= 2.4L

T1 = 0°C + 273 K = 273 K

T2 = 100°C + 273 K = 373 K

We have to find the pressure of the gas, by using the gas formula as,

$\frac{P 1 V 1}{T 1}=\frac{P 2 V 2}{T 2}

P2 can be found by rewriting the above expression as,

$P 2=\frac{P 1 \times V 1 \times T 2}{T 1 \times V 2}

Plugin the above values as,

$P 2=\frac{1 \text {atm} \times 10 L \times 373 \mathrm{K}}{273 \mathrm{K} \times 2.4 \mathrm{L}}=5.7 \text { atm }

4 0
2 years ago
Which statements best describe half-lives of radioactive isotopes? Check all that apply.
Igoryamba

Answer:

The half-life varies depending on the isotope.

Half-lives range from fractions of a second to billions of years.

The half-life of a particular isotope is constant.

3 0
2 years ago
Read 2 more answers
a.) The diameter of a uranium atom is 3.50 Å . Express the radius of a uranium atom in both meters and nanometers. b.) How many
Svetradugi [14.3K]

Answer:

Th answer to your question is:

a)  3.5 x10⁻¹⁰ meters; 0.35 nm

b) 6857142.86 atoms

c) Volume = 2.06 x 10⁻²³ cm³

Explanation:

a) data

Uranium atoms = 3.5A°

meters

           1 A° ----------------  1 x 10 ⁻¹⁰ m

         3.5A° ---------------  x

 x = 3.5(1 x10⁻¹⁰)/ 1 = 3.5 x10⁻¹⁰ meters

          1 A° ------------------ 0.1 nm

        3.5 A° ---------------- 0.35 nm

b) 2.4 mm

Divide 2,40 mm / uranium diameter

But, first convert 3,5A° to mm   = 3.5 x 10⁻⁷ mm

# of uranium atoms = 2.4 / 3.5 x 10⁻⁷ = 6857142.86

c) volume in cubic cm

Convert 3.5A° to cm  = 3.5 x 10⁻⁸

Volume = 4/3 πr³ = (4/3) (3.14)(1.7 x10⁻⁸)³

Volume = 2.06 x 10⁻²³ cm³

6 0
2 years ago
How many moles of al2o3 can be produced from the reaction of 10.0 g of al and 19.0 of o2?
Advocard [28]

Answer:

0.185moles of Al₂O₃

Explanation:

Mass of Al = 10g

Mass of O₂ = 19g

Equation of the reaction: 4Al + 3O₂ → 2Al₂O₃

This is the balanced reaction equation.

Solution

From the given parameters, the reactant that would determine the extent of the reaction is Aluminium. It is called the limiting reagent. Oxygen is in excess and it is in an unlimited supply.

Working from the known mass to the unknown, we simply solve for the number of moles of Al using the mass given.

Then from the equation, we can relate the number of moles of Al to that of Al₂O₃ produced:

 Number of moles of Al = \frac{mass}{molar mass}

                                        =   \frac{10}{27}

                                        = 0.37mol

From the equation:

         4 moles of Al produced 2 moles of Al₂O₃

    0.37 mole will yield:  \frac{2 x 0.37}{4} = 0.185moles of Al₂O₃

8 0
2 years ago
For the following dehydrohalogenation (E2) reaction, draw the Zaitsev product(s) resulting from elimination involving C3–C4 (i.e
Bond [772]

Answer:

2-methyl-butene

Explanation:

For the E2 mechanism, we have an <u>anti-elimination</u>. The Br leaves the molecule and the base removes the hydrogen in the anti position to form the double that's why only one structure is produced. (See figure 1)

Since we have 2 hydrogens on the right carbon, we cannot indicate a <u>specific stereoisomer</u>, in other words, it is not possible to assign a <u>Z or E</u> configuration for this alkene.

8 0
2 years ago
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