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Tema [17]
2 years ago
14

Liquid nitrogen has a density of 0.807 g/ml at –195.8 °c. if 1.00 l of n2(l) is allowed to warm to 25°c at a pressure of 1.00 at

m, what volume will the gas occupy? (r = 0.08206 l×atm/k×mol)
Chemistry
1 answer:
Flauer [41]2 years ago
8 0
Step 1: Change density from g/mL to g/L;

                                  0.807 g/mL  =  807 g/L

Step 2: Find Moles of N₂;
As,
             Density  =  Mass / Volume
Or,
             Mass  =  Density × Volume

Putting Values,

            Mass  =  807 g/L × 1 L

            Mass  =  807 g
Also,
            Moles  =  Mass / M.mass

Putting values,

            Moles  =  807 g / 28 g.mol⁻¹

            Moles  =  28.82 moles

Step 3: Apply Ideal Gas Equation to Find Volume of gas occupied,

As,
                        P V  =  n R T

                            V  =  n R T / P
Putting Values, remember! don't forget to change temperatue into Kelvin (25 °C + 273 = 298 K)

                 V  =  (28.82 mol × 0.08206 atm.L.mol⁻¹.K⁻¹ × 298 K) ÷ 1 atm

                 V  =  704.76 L
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Check 0/1 ptRetries 5 Element R has three isotopes. The isotopes are present in 0.0825, 0.2671, and 0.6504 relative abundance. I
Tcecarenko [31]

Answer:

Atomic mass = 127.198 amu

Explanation:

The average atomic mass is obtained by summing the masses of the isotopes each multiplied by its abundance.

Atomic mass = (97.62 * 0.0825) + (109.3 * 0.2671) + (138.3 * 0.6504)

Atomic mass = 8.05365 +  29.19403 + 89.95032

Atomic mass = 127.198 amu

7 0
2 years ago
Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by No
liraira [26]

Answer:

  • Molality = 0.066 m
  • Molar mass = 608.36 g/mol

Explanation:

It seems the question is incomplete. However a web search us shows this data:

" Reserpine is a natural product isolated from the roots of the shrub Rauwolfia serpentina. It was first synthesized in 1956 by Nobel Prize winner R. B. Woodward. It is used as a tranquilizer and sedative. When 1.00 g reserpine is dissolved in 25.0 g camphor, the freezing-point depression is 2.63 °C (Kf for camphor is 40 °C·kg/mol). Calculate the molality of the solution and the molar mass of reserpine. "

The <em>freezing-point depression</em> is expressed by:

  • ΔT=Kf * m

We put the data given by the problem and <u>solve for m</u>:

  • 2.63 °C = 40°C·kg/mol * m
  • m = 0.06575 m

For the calculation of the molar mass:<em> Molality</em> is defined as moles of solute per kilogram of solvent:

  • 0.06575 m = Moles reserpine / kg camphor
  • 25.0 g camphor ⇒ 25.0/1000 = 0.025 kg camphor

We<u> calculate moles of reserpine:</u>

  • 0.06575 m = Moles reserpine / 0.025 kg camphor
  • Moles reserpine = 1.64x10⁻³ mol

Finally we use the mass of reserpine and the moles to calculate <u>the molar mass</u>:

  • 1.00 g reserpine / 1.64x10⁻³ mol = 608.36 g/mol

<em>Keep in mind that if the data in your problem is different, the results will be different. But the solving method remains the same.</em>

8 0
2 years ago
How many grams of the excess reagent are left over when 6.00g of CS2 gas react with 10.0g of Cl2 gas in the following reaction:
WARRIOR [948]

Answer:

CS₂ = 2.43 g

Explanation:

Data Given:

CS₂ gas = 6.00g

Cl₂ = 10.0g

Reaction Given:

                   CS₂(g) + 3Cl₂(g) --------> CCl₄(l) + S₂Cl₂(l)

Solution

Limiting Reagent :

The reactant which is less in amount control the amount of product and know as limiting reagent.

Excess Reagent:

The amount of reactant which are in excess and leftover at the end of reaction and product formed.

Now we have to find the reactant that is in excess

For this we will look at the Reaction

                       CS₂   +    3Cl₂  -------->   CCl₄   +  S₂Cl₂

                       1 mol     3 mol               1 mol      1 mol

we come to know from the above reaction that

1 mole of CS₂ react with 3 mole of Cl₂ to produce 1 mole of CCl₄ and 1 mole of  S₂Cl₂

Now to convert mole to mass we required molar masses

molar mass of CS₂ = 12 +  2(32)

molar mass of CS₂ = 76 g/mol

molar mass of Cl₂ = 2(35.5)

molar mass of Cl₂ = 71 g/mol

if we represent mole in grams then

             CS₂              +    3Cl₂             -------->   CCl₄   +  S₂Cl₂

      1 mol (76 g/mol)     3 mol (71 g/mol)

                CS₂   +    3Cl₂   -------->   CCl₄   +  S₂Cl₂

                 76 g       213 g

So,

we come to know that 76 g of CS₂ will combine with 213 g of Cl₂ to form product.

So now look for the ratio of both reactant

                   CS₂    :    Cl₂

                   76 g   :    213 g

                    1  g     :      2.8 g

So, the for every one gram of CS₂ required 2.8g Cl₂

So from this details

we apply unity formula

                 2.8 g of Cl₂ ≅  1 g of CS₂

                 10 g of Cl₂ ≅  ? g of CS₂

by doing cross multiplication

                    g of CS₂ = 10 x 1 / 2.8

                    g of CS₂ = 3.6 g

So,

10.0g of Cl₂ used 3.57 g of CS₂

It showed that 3.57 g of CS₂ used and the remaining amount of CS₂ is

               Remaining amount of CS₂ = 6 - 3.57 = 2.43 g

 

8 0
2 years ago
In science class, Blaine’s teacher puts one glow stick in a cup of hot water and another glow stick in a cup of cold water. She
timofeeve [1]

Answer:

The glow stick in hot water will be brighter

Explanation:

The glow stick in hot water will be brighter than the glow stick in cold water because the heat from the hot water will cause the molecules in the glow stick to move faster. The faster the molecules move in the glow stick, the sooner and brighter the reaction will be. The cold water will cause molecules to move slowly and it will take longer for the reaction to occur, which will also make it less bright.

3 0
2 years ago
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A crystal of graphite contains which type of bond?
cupoosta [38]

Answer:Graphite has a giant covalent structure in which: each carbon atom is joined to three other carbon atoms by covalent bonds.

4 0
2 years ago
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