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4 years ago

Why do you think a building material's density is important when building homes or architectural structures?

Chemistry

2 answers:

guapka [62]4 years ago

8 0

The Density Is Important For Making Sure It's Steady and Firm

Fynjy0 [20]4 years ago

4 0

Answer: In order to maintain the stability, the materials that are used for constructing buildings and other structures must be dense enough. This materials are such as brick, woods, steel. The more denser it is, the more stable it is.

Buildings that are made up of lesser density materials results in the collapsing or subsidence, during earthquake. So the materials must be carefully chosen and it is an important factor that is to be kept in mind by the engineering geologist.

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The equation represents the decomposition of a generic diatomic element in its standard state. 12X2(g)⟶X(g) Assume that the stan

aliina [53]

Answer:

$K^{2000K}=0.774\\\\K^{3000K}=12.56$

Explanation:

Hello,

In this case, considering the reaction, we can compute the Gibbs free energy of reaction at each temperature, taking into account that the Gibbs free energy for the diatomic element is 0 kJ/mol:

$\Delta _rG=\Delta _fG_{X}-\frac{1}{2} \Delta _fG_{X_2}=\Delta _fG_{X}$

Thus, at 2000 K:

$\Delta _rG=\Delta _fG_{X}^{2000K}=4.25kJ/mol$

And at 3000 K:

$\Delta _rG=\Delta _fG_{X}^{3000K}=-63.12kJ/mol$

Next, since the relationship between the equilibrium constant and the Gibbs free energy of reaction is:

$K=exp(-\frac{\Delta _rG}{RT} )$

Thus, at each temperature we obtain:

$K^{2000K}=exp(-\frac{4250J/mol}{8.314\frac{J}{mol\times K}*2000K} )=0.774\\\\K^{3000K}=exp(-\frac{-63120J/mol}{8.314\frac{J}{mol\times K}*3000K} )=12.56$

In such a way, we can also conclude that at 2000 K reaction is unfavorable (K<1) and at 3000 K reaction is favorable (K>1).

Best regards.

4 0

3 years ago

Which of the following gases will have the slowest rate of effusion?

Neko [114]

hydrogen gas...........

6 0

4 years ago

For a particular reaction, ΔH∘=20.1 kJ/mol and Δ????∘=45.9 J/(mol⋅K). Assuming these values change very little with temperature,

tia_tia [17]

Answer:

The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous

Explanation:

<u>Step 1:</u> Data given

ΔH∘=20.1 kJ/mol

ΔS is 45.9 J/K

<u>Step 2:</u> When is the reaction spontaneous

Consider temperature and pressure = constant.

The conditions for spontaneous reactions are:

ΔH <0

ΔS > 0

ΔG <0 The reaction is spontaneous at all temperatures

ΔH <0

ΔS <0

ΔG <0 The reaction is spontaneous at low temperatures ( ΔH - T*ΔS <0)

ΔH >0

ΔS >0

ΔG <0 The reaction is spontaneous at high temperatures ( ΔH - T*ΔS <0)

<u>Step 3:</u> Calculate the temperature

ΔG <0 = ΔH - T*ΔS

T*ΔS > ΔH

T > ΔH/ΔS

In this situation:

T > (20100 J)/(45.9 J/K)

T > 437.9 K

T > 164.75 °C

The temperature should be higher than 437.9 Kelvin (or 164.75 °C) to be spontaneous

7 0

3 years ago

What is the mass of a 3.5 l sample of a liquid that has a density of 1.2 g/ml?

Aleks [24]

Hey there!:

Volume in mL:

3.5 * 1000 = 3500 mL

density = 1.2 g/mL

Therefore:

D = m / V

1.2 = m / 3.5

m = 1.2 * 3.5

m = 4.2 g

hope this helps!

7 0

3 years ago

For the reaction Halg) + 12/8) = 2HI(g), K = 50.2 at 445°C. If

artcher [175]

C) Qc < Kc, the reaction proceeds from left to right to reach equilibrium

<h3>Further explanation</h3>

Given

K = 50.2 at 445°C

[H2] = [I2] = [HI] = 1.75 × 10⁻³ M At 445ºC

Reaction

H2(g) + I2(g) ⇔2HI(g)

Required

Solution

Qc for the reaction

$\tt Qc=\dfrac{[HI]^2}{[H_2][I_2]}\\\\Qc=\dfrac{(1.75.10^{-3})^2}{1.75.10^{-3}\times 1.7\times 10^{-3}}=1$

Qc < Kc ⇒ reaction from left(reactants) to right (products) (the reaction will shift on the right) until it reaches equilibrium (Qc = Kc)

5 0

3 years ago