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2 years ago

5. An object is a regular, rectangular, solid with dimensions of 2 cm by 3cm by 2cm. It has a mass of 24 g. find its density.

Chemistry

1 answer:

m_a_m_a [10]2 years ago

3 0

Answer:

Density, D = 2g/cm^3

Explanation:

Given the following data;

Length = 2cm

Width = 3cm

Height = 2cm

Mass = 24g

Density = ?

Volume of a rectangular solid (V) = Length × Weight × Height

Therefore, V = L× W × H

Substituting the values, we have;

$V = 2 *3 * 2 = 12$

V = 12cm

Density can be defined as the ratio of mass to volume i.e mass all over volume.

Mathematically, $Density = \frac{Mass}{Volume}$

$D = \frac{M}{V}$

Substituting the values, we have;

$D = \frac{24}{12}$

<em>Density, D = 2g/cm^3</em>

Hence, the density of the rectangular solid is 2g/cm^3.

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If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________. If the initial concen

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This is an incomplete question, here is a complete question.

If the initial concentration of NOBr is 0.0440 M, the concentration of NOBr after 9.0 seconds is ________.

The reaction

$2NOBr(g)\rightarrow 2NO(g)+Br_2(g)$

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Answer : The concentration of after 9.0 seconds is, 0.00734 M

Explanation :

The integrated rate law equation for second order reaction follows:

$k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)$

where,

k = rate constant = 0.80 M⁻¹s⁻¹

t = time taken = 142 second

[A] = concentration of substance after time 't' = ?

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Putting values in above equation, we get:

$0.80M^{-1}s^{-1}=\frac{1}{142s}\left (\frac{1}{[A]}-\frac{1}{(0.0440M)}\right)$

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1 year ago

Explain why the spectrum produced by a 1-gram sample of element Z would have the same spectral lines at the same wavelengths as

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2 years ago

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A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

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In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

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<h3>Answer:</h3>

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