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3 years ago

Briefly describe four ways that a beam can be strengthened?

Chemistry

1 answer:

lara31 [8.8K]3 years ago

8 0

Answer:

1. Changing Beam Material

2. Corrugation

3. Changing Beam form

4. Steel Reinforcing Bars

Explanation:

Changing Beam Material

Some materials are stronger when used in beams than others. Beams made of steel for instance are stronger than beams made of wood. Therefore changing material can improve the strength of the beam. It is quite important to take into account the weights of the material though as different structures have different requirements.

Corrugation.

You can fold the beam into triangular shapes to increase strength. If you look at roofs you will notice that they are folded and this increased their strength. The same logic can be applied to beams.

Changing Beam Form

Another way to make Beams stronger is to change their form or rather their shape. Straight beams are not as strong as I-beams for instance. I-beams look like the capital letter I with the lines at both ends. I-beams are usually used in construction which shows that they are quite strong.

Steel Reinforcing Bars

When placed in concrete beams, Steel Reinforcing Bars which are also called Rebar can help strengthen a beam by helping it withstand the forces of tension. A concrete beam with Rebar inside it is known as Reinforced Concrete.

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What is the calculated value of the cell potential at 298K for an

Tpy6a [65]

The question is incomplete, here is the complete question:

What is the calculated value of the cell potential at 298 K for an electrochemical cell with the following reaction, when the H₂ pressure is 6.56 x 10⁻² atm, the H⁺ concentration is 1.39 M, and the Sn²⁺ concentration is 9.35 x 10⁻⁴ M?

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

Answer: The cell potential of the given electrochemical cell is 0.273 V

Explanation:

For the given chemical equation:

$2H^+(aq)+Sn(s)\rightarrow H_2(g)+Sn^{2+}(aq)$

The half cell reactions for the given equation follows:

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $H_2+2e^-\rightarrow H_2(g);E^o_{2H^{+}/H_2}=0.0V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=0.0-(-0.14)=0.14V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]\times p_{H_2}}{[H^+]^2}$

where,

$E_{cell}$ = electrode potential of the cell = ?

$E^o_{cell}$ = standard electrode potential of the cell = +0.14 V

n = number of electrons exchanged = 2

$[H^{+}]=1.39M$

$[Sn^{2+}]=9.35\times 10^{-4}M$

$p_{H_2}=6.56\times 10^{-2}atm$

Putting values in above equation, we get:

$E_{cell}=0.14-\frac{0.059}{2}\times \log(\frac{(9.35\times 10^{-4})\times (6.56\times 10^{-2})}{(1.39)^2})\\\\E_{cell}=0.273V$

Hence, the cell potential of the given electrochemical cell is 0.273 V

4 0

3 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:

Step 1: Data given

ΔH∘=20.1 kJ/mol

ΔS is 45.9 J/K

Step 2: 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)

Step 3: 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

For the balanced chemical reaction

musickatia [10]

Answer:

150

Explanation:

C₄H₂OH + 6O2 → 4CO2 + 5H₂O

We can find the equivalent number of O₂ molecules for 100 molecules of CO₂ using a conversion factor containing the stoichiometric coefficients of the balanced reaction, as follows:

100 molecules CO₂ * $\frac{6moleculesO_2}{4moleculesCO_2}$ = 150 molecules O₂

150 molecules of O₂ would produce 100 molecules of CO₂.

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Answer:

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Briefly describe four ways that a beam can be strengthened?​

Briefly describe four ways that a beam can be strengthened?