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3 years ago

What is the percent yield if the theoretical yield of ChCl3 is 14.50 g but the actual yield is 12.33 g?

1 answer:

4 0

Percentage is a ration of part to a whole. In our case, in perfect conditions, yield is theoretical and thusly considered larger, a whole. While actual yield is less because of imperfect conditions. The percentage is,

$(12.33\cdot 100)/14.5\approx85.03\%$ .

Hope this helps.

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3 years ago

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When a piece of metal is irradiated with UV radiation (λ = 162 nm), electrons are ejected with a kinetic energy of 3.54×10-19 J.

dsp73

We have that the work function of the metal

$\phi=1.227*10^{-18}J$

From the Question we are told that

UV radiation (λ = 162 nm)

Kinetic energy $K.E =3.54*10-19 J.$

Generally the equation for Kinetic energy is mathematically given as

$KE =\frac{hc}{\pi-\phi} \\\\\phi =\frac{ 6.626*10^{-34} * 3*10^8}{162*10^{-9} -3.54*10^{-19}}$

$\phi=1.227*10^{-18}J$

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3 years ago

Explain the role of a primary consumer and a secondary consumer in a food web.

Usimov [2.4K]

Answer:

Explanation:

Secondary consumers are organisms that eat primary consumers for energy. Primary consumers are always herbivores, or organisms that only eat autotrophic plants.

Carnivores only eat other animals, and omnivores eat both plant and animal matter.

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3 years ago

The equilibrium constant Kc for the reaction PCl3(g) + Cl2(g) ⇌ PCl5(g) is 49 at 230°C. If 0.70 mol of PCl3 is added to 0.70 mol

Ymorist [56]

Answer : The correct option is, (B) 0.11 M

Solution :

First we have to calculate the concentration $PCl_3$ and $Cl_2$ .

$\text{Concentration of }PCl_3=\frac{\text{Moles of }PCl_3}{\text{Volume of solution}}$

$\text{Concentration of }PCl_3=\frac{0.70moles}{1.0L}=0.70M$

$\text{Concentration of }Cl_2=\frac{\text{Moles of }Cl_2}{\text{Volume of solution}}$

$\text{Concentration of }Cl_2=\frac{0.70moles}{1.0L}=0.70M$

The given equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

Initially 0.70 0.70 0

At equilibrium (0.70-x) (0.70-x) x

The expression of $K_c$ will be,

$K_c=\frac{[PCl_5]}{[PCl_3][Cl_2]}$

$K_c=\frac{(x)}{(0.70-x)\times (0.70-x)}$

Now put all the given values in the above expression, we get:

$49=\frac{(x)}{(0.70-x)\times (0.70-x)}$

By solving the term x, we get

$x=0.59\text{ and }0.83$

From the values of 'x' we conclude that, x = 0.83 can not more than initial concentration. So, the value of 'x' which is equal to 0.83 is not consider.

Thus, the concentration of $PCl_3$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $Cl_2$ at equilibrium = (0.70-x) = (0.70-0.59) = 0.11 M

The concentration of $PCl_5$ at equilibrium = x = 0.59 M

Therefore, the concentration of $PCl_3$ at equilibrium is 0.11 M

3 0

3 years ago

An organic synthesis to make the pain reliever acetaminophen is supposed to produce 280 kg of product but instead produces 70 kg

Irina-Kira [14]

Answer: 75%

Explanation:

The following information can be gotten from the question:

Waste = 70kg

Theoretical yield = 280kg

Therefore, the actual yield will be the difference between the theoretical yield and the waste which will be:

= 280kg - 70kg = 210kg

The percent yield will now be:

= Actual yield / Theoretical yield × 100

= 210/280 × 100

= 3/4 × 100

= 75%

4 0

3 years ago