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Rainbow [258]
3 years ago
10

When ethane (c2h6) reacts with chlorine (cl2) the main product is c2h5cl; but other products containing cl (chlorine), such as c

2h4cl2, are also obtained in small quantities. the formation of these other products reduces the yield of c2h5cl. in a certain experiment 145 g of c2h6 reacts with 215 g of cl2. (a) assuming that c2h6 and cl2 react only to form c2h5cl and hcl, calculate the theoretical yield of c2h5cl?
Physics
1 answer:
Deffense [45]3 years ago
5 0
The balanced chemical reaction is written as:

C₂H₆ + Cl₂ → C₂H₅Cl + HCl

Mol of C₂H₆ available: 145 g/ 30 g/mol = 4.833 mol
Mol of Cl₂ available: 215 g/ 70.9 g/mol = 3.032 mol

So, the limiting reactant is Cl₂. We base our theoretical yield from here.

3.032 mol Cl₂ (1 mol C₂H₅Cl/ 1 mol Cl₂)(64.45 g/mol C₂H₅Cl) = <em>195.41 g C₂H₅Cl</em>


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In a Young’s interference experiment, the two slits are separated by 0.150 mm and the incident light includes two wavelengths: l
dezoksy [38]

Answer:

2.52 × 10⁻² cm

Explanation:

The distance of bright fringe from the center of the screen is given by the formula

                y = \frac{m\lambda D}{d}

Here, wavelength is λ, Distance of the screen from the slits is D, seperation between the

slits is d.

   

Separation between the slits, d = 0.15 mm

                                                     = 0.15 * 10^{-3} m

Distance of the screen from the slits = 1.40 m

We have a wavelength, λ1 = 540 nm

                                           = 540 * 10^{-9} m

By substituting all these values in the above equation we get

                y1 = mλD/d

                y1 = m(540 \times 10^{-9} m)(1.40 m)/(0.15 \times 10^{-3} m)

                y1 = m(5.04 * 10^{-3} m)

We have a wavelength, λ2 = 450 nm

                                           = 450 * 10^-9 m

By substituting all these values in the above equation we get

                y_2 = \frac{m\lambda D}{d}

                y_2 = m(450 * 10^{-9} m)(1.40 m)/(0.15 * 10^{-3} m)

                y_2 = m'(4.20 * 10^{-3} m)

According to the problem, these two distance are coincides with each other.

So,

                           y_1 = y_2

m(5.04 * 10^{-3} m) = m'(4.20 * 10^{-3} m)

by testing values, the above equation is satisfied only when, m = 5 and m' = 6

Then from the above we have

                           y1 = y2 = 0.0252 m

                                         = 2.52 × 10⁻² cm

8 0
3 years ago
Which of the following is an example of a SOLUTION?
Mila [183]

Answer:

Salt water

Explanation:

This is the answer

7 0
2 years ago
A 0.15 kg mass is suspended from a vertical spring and descends a distance of 4.6 cm, after which it hangs at rest. An additiona
ch4aika [34]

Answer:

The total extension = 19.9 cm

Work done in stretching the spring through both displacement = <em>0.633 J.</em>

Explanation:

From Hook's law,

F  = ke .................................... Equation 1.

Where F = force or weight, k = force constant of the spring, e = extension.

making k the subject of the equation,

k = F/e .......................... Equation 2

Given: F = W = mg = 0.15(9.8) = 1.47 N, e = 4.6 cm = 0.046 m.

k = 1.47/0.046

k = 31.96 N/m.

When an additional mass of 0.5 kg is suspended,

Total mass suspended(M₁) = 0.15+0.5 = 0.65 kg

Total Weight (W₁) = 0.65(9.8) = 6.37 N.

k = 31.96 N/m

Substituting into equation 1

6.37 = 31.96e

e = 6.37/31.96

e = 0.199 m

e = 19.9 cm

Thus the total extension = 19.9 cm

a.

Work done in stretching the spring  through both displacement

W = 1/2ke²..................... Equation 3

Where W = work done, k = spring constant, e = extension.

Given: k = 31.96 N/m, e = 0.199 m

Substituting into equation 3

W = 1/2(31.96)(0.199)²

<em>W = 0.633 J.</em>

5 0
3 years ago
(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t)
dolphi86 [110]

Answer:

r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k

Explanation:

a(t)=20t i+sin(t) j +cos(2t) k

v(t)=\int\limits^a_b {a(t)} \, dt

=(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k---------------eqn 1

given v(0)=i

i=c_1i+(-1+c_2)j+(0+c_3)k

c_1=1   c_2=1  c_3=0

from equation 1

V(t)=(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k----------eqn 2

now r(t)=\int\limits^a_b {v (t)} \, dt

(\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k

given r(0)=j

0i+1j+ok = c_1i+c_2j+(\frac{-1}{4}+c_3)k

c_1=0  c_2=0  c_3 =  \frac{1}{4}

r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k

8 0
3 years ago
Which two options are forms of potential energy
blsea [12.9K]

Answer:

e and b

Explanation:

6 0
3 years ago
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