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3 years ago

State the career function of chemical engineering

Physics

2 answers:

rusak2 [61]3 years ago

8 0

Answer:

The main role of chemical engineers is to design and troubleshoot processes for the production of chemicals, fuels, foods, pharmaceuticals, and biologicals, just to name a few. They are most often employed by large-scale manufacturing plants to maximize productivity and product quality while minimizing costs

Explanation:

Arte-miy333 [17]3 years ago

7 0

Answer:

Chemical engineers develop and design chemical manufacturing processes. They apply the principles of chemistry, biology, physics, and math to answer problems that involve the production of chemicals, fuel, drugs, food, and many products

Explanation:

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Hydrogen fluoride gas (HF) and sodium hydroxide (NaOH) react in a test tube. They form water and sodium fluoride (NaF). Which ty

AfilCa [17]

This reaction is a double replacement reaction. Hydrogen in the hydrogen fluoride gets replaced by sodium of the sodium hydroxide. HF is a weak acid and NaOH is a strong base. So, double replacement produces sodium fluoride and water molecules.

6 0

3 years ago

Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse

777dan777 [17]

Answer:

Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

d sin θ = (m + ½) λ

As the observation is on a distant screen

tan θ = y / x

tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

tan θ = sin θ

Let's replace

d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance

y₁ = (m + ½) λ x / d

m = 1

y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

m = 2

y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

Δy = λ x / d (5/2 -3/2)

Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

Δy= 5,075 10⁻⁶ m

8 0

3 years ago

A dinner plate falls vertically to the floor and breaks up into three pieces, which slide horizontally along the floor. Immediat

N76 [4]

Answer:

p_k=\sqrt{p_x^2+p_y^2}}

Explanation:

Apply the momentum in each direction knowing that the impact is at the same time for the pieces so

$p_x=m_1*v_1$

$p_x=200g*2.0m/s=0.4kgm/s$

$p_y=m_2*v_2$

$p_y=235g*1.5m/s=0.3525kgm/s$

So the momentum in the other piece can be find knowing that

$p_x^2+p_y^2=p_k^2$

So:

$p_k=\sqrt{p_x^2+p_y^2}}$

$p_k=\sqrt{0.4^2+0.3525^2}}=\sqrt{0.2842 kg^2*m^2/s^2}$

$p_k=0.5331kg*m/s$

To find the velocity knowing the mass

$p_k=m_k*v_k$

$v_k=\frac{p_k}{m_k}=\frac{0.5331 kg*m/s}{0.10kg}$

$v_k=5.331 m/s$

4 0

3 years ago

Calculate the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal. Assume that the spe

goblinko [34]

Answer:

the ratio of the kinetic energy of an electron to that of a proton if their wavelengths are equal is 1835.16 .

Explanation:

We know, wavelength is expressed in terms of Kinetic Energy by :

$\lambda=\dfrac{h}{\sqrt{2mE}}$

Therefore , $E=\dfrac{h^2}{2 \lambda^2 m}$

It is given that both electron and proton have same wavelength.

Therefore,

$E_e=\dfrac{h^2}{2 \lambda^2 m_e}$ .... equation 1.

$E_p=\dfrac{h^2}{2 \lambda^2 m_p}$ .... equation 2.

Now, dividing equation 1 by 2 .

We get ,

$\dfrac{E_e}{E_p}=\dfrac{\dfrac{h^2}{2 \lambda^2 m_e}}{\dfrac{h^2}{2 \lambda^2 m_p}}\\\\\\\dfrac{E_e}{E_p}=\dfrac{m_p}{m_e}$

Putting value of mass of electron = $9.1\times 10^{-31}\ kg$ and mass of proton = $1.67\times 10^{-27}\ kg.$

We get :

$\dfrac{E_e}{E_p}=\dfrac{1.67\times 10^{-27}\ kg}{9.1\times 10^{-31}\ kg}=1835.16$

Hence , this is the required solution.

4 0

4 years ago

Read 2 more answers

Air having a pressure of 40 psig and a volume of 8 cu ft expands isothermally to a pressure of 10 psig. Find the external work p

DerKrebs [107]

Answer:

357.6 lb-ft

Explanation:

V = Volume = 8 ft³

dP = Change in pressure = (40-10) = 30 psig

Work done is given by

$W=VdP\\\Rightarrow W=8\times (40-10)\\\Rightarrow W=240\ psig.ft^3$

$30\ psig=44.7\ psi\\\Rightarrow 1\ psi=\dfrac{30}{44.7}$

So, converting to ft-lb

$\dfrac{240}{\dfrac{30}{44.7}}=357.6\ lb-ft$

The external work performed during the expansion is 357.6 lb-ft

7 0

4 years ago

State the career function of chemical engineering​

State the career function of chemical engineering