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Ksju [112]
3 years ago
13

Using mathematical expressions drive the compression factor​

Chemistry
1 answer:
Zarrin [17]3 years ago
6 0

Answer:

just look it up yourself

Explanation:

ITS NOT THAT HARD

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Show the calculation of the mass (grams) of a 14.5 liter gas sample with a molecular weight of 82 when collected at 29°C and 740
mote1985 [20]

Answer:

In this conditions, the gaswll weight 46.74 g.

Explanation:

The idal gas law states that:

PV = nRT,

P: pressure = 740 mmHg = 0.97 atm

V: volume = 14.5 L

n: number of moles

R: gas constant =0.08205 L.atm/mol.K

T: temperature = 29°C = 302.15K

n = \frac{PV}{RT} \\n = \frac{0.97x14.5}{0.082 x 302.15 } \\n = 0.57 mol

1 mol gas ___ 82 g

0.57 mol gas __ x

x = 46.74 g

4 0
3 years ago
What element was named for the scientist who discovered the nucleus of the atom using gold foil
Artist 52 [7]
<span>rutherfordium element # 104</span>
4 0
3 years ago
Calculate the electric double layer thickness of a alumina colloid in a dilute (0.1 mol/dm3) CsCI electrolyte solution at 30 °C.
Ad libitum [116K]

Explanation:

The given data is as follows.

    Concentration = 0.1 mol/dm^{3}

                             = 0.1 \frac{mol dm^{3}}{dm^{3}} \frac{10^{3}}{dm^{3}} \times \frac{6.022 \times 10^{23}}{1 mol} ions

                             = 6.022 \times 10^{25} ions/m^{3}

               T = 30^{o}C = (30 + 273) K = 303 K

Formula for electric double layer thickness (\lambda_{D}) is as follows.

            \lambda_{D} = \frac{1}{k} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}

where, n^{o} = concentration = 6.022 \times 10^{25} ions/m^{3}

Hence, putting the given values into the above equation as follows.

                 \lambda_{D} = \sqrt \frac{\varepsilon \varepsilon_{o} K_{g}T}{2 n^{o} z^{2} \varepsilon^{2}}                    

                          = \sqrt \frac{78 \times 8.854 \times 10^{-12} c^{2}/Jm \times 1.38 \times 10^{-23}J/K \times 303 K}{2 \times 6.022 \times 10^{25} ions/m^{3} \times (1)^{2} \times (1.6 \times 10^{-19}C)^{2}}  

                         = 9.669 \times 10^{-10} m

or,                     = 9.7 A^{o}

                          = 1 nm (approx)

Also, it is known that \lambda_{D} = \sqrt \frac{1}{n^{o}}

Hence, we can conclude that addition of 0.1 mol/dm^{3} of KCl in 0.1 mol/dm^{3} of NaBr "\lambda_{D}" will decrease but not significantly.

7 0
3 years ago
How to put Roman numerals in a science equation
Bogdan [553]
Example:

Oxygen(III)
6 0
3 years ago
PLEASE HELP!! EXPLAIN YOUR ANSWER PLEASE! :)
GaryK [48]
The formula for force is F=ma. Because weight is a measure of force, then we can substitute the weight of the meteor, 3204 N, for F in the the equation for force. We also know that the acceleration of gravity on Earth is 9.8 m/s^2. To find the mass, simply divide both sides of the equation by the value of acceleration to get
m =   \frac{f}{a}  =  \frac{3204}{9.8}  = 326.9
Therefore, the value of the mass of the meteor is 326.9 kg.
8 0
3 years ago
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