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Tema [17]
3 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]3 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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A block of metal has a volume of 14.0in3 and weighs 5.16 lb .
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The density of metal block in grams per cubic centimeter is 10.70 g/cm³.

Given,

Mass of metal block = 5.16 lb  

1 lb = 453.592 g

5.26 lb = 2340.536 g

The volume of metal block = 14 in 3

1 in = 2.5 cm

1 in 3 = 15.625 cm³

14 in 3 = 218.75 cm³

Density is defined as the mass per unit volume of a substance. Or, it is the ratio of mass to the volume of the substance.

As we know,

Density = mass/volume

Or, density = 2340.536 / 218.75

Or, density = 10.70 g/cm³

Therefore, the density of the metal block is 10.70 g/cm³.

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Protons is the answer
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A student prepares a solution of Potassium Nitrate (KNO3) containing 95g at 40 C. This solution is -
Sholpan [36]

Answer:

unsaturated

Explanation:

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You are given an unknown type of clothing dye. how could you use the procedures in this lab to see if this dye is a mixture?
Arte-miy333 [17]
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A student dissolved 1.805g of a monoacidic weak base in 55mL of water. Calculate the equilibrium pH for the weak monoacidic base
yawa3891 [41]

Answer:

11.39

Explanation:

Given that:

pK_{b}=4.82

K_{b}=10^{-4.82}=1.5136\times 10^{-5}

Given that:

Mass = 1.805 g

Molar mass = 82.0343 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{1.805\ g}{82.0343\ g/mol}

Moles= 0.022\ moles

Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)

Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}

Molarity=\frac{0.022}{0.055}

Concentration = 0.4 M

Consider the ICE take for the dissociation of the base as:

                                  B +   H₂O    ⇄     BH⁺ +        OH⁻

At t=0                        0.4                          -              -

At t =equilibrium     (0.4-x)                        x           x            

The expression for dissociation constant is:

K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}

1.5136\times 10^{-5}=\frac {x^2}{0.4-x}

x is very small, so (0.4 - x) ≅ 0.4

Solving for x, we get:

x = 2.4606×10⁻³  M

pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61

<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>

5 0
3 years ago
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