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Mamont248 [21]
3 years ago
6

What is the formula used to determine the % yield?

Chemistry
1 answer:
Illusion [34]3 years ago
8 0
You multiply the percentage and theoretical yield together.
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Why is your weight less on the Moon than on Earth, but your mass is the same?​
Anon25 [30]

Answer:

Mass is the amount of matter in the object, and is not affected by gravity. Weight, on the other hand, is directly related to gravity and how much you pull to the ground.

8 0
4 years ago
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A glass flask of volume 400cm is just filled with mercury at 0 degree ​
3241004551 [841]

<em>A glass flask of volume 400 cm³ is just filled with mercury at 0°C. How much mercury will  overflow when the temperature of the system rises to 80°C.​</em>

<em />

The volume of mercury that overflow is 5.376 cm³

<h3>Further explanation</h3>

Given

volume of glass = 400 cm³

Δt=80 °C - 0 °C = 80

Required

overflow volume

Solution

With an increase in the temperature of the substance, objects can expand. This expansion includes volume expansion.

Can be formulated

\tt V_t=V_o(1+\gamma.\Delta t)\\\\\Delta V=V_o.\gamma.\Delta t

Find volume expansion of glass and mercury

  • Glass(γ = 1.2.10⁻⁵ C⁻¹)

\tt \Delta V=400\times 1.2.10^{-5}.80=0.384~cm^3

  • Mercury(γ=1.8.10⁻⁴ C⁻¹)

\tt \Delta V=400\times 1.8.10^{-4}\times 80=5.76~cm^3

Overflow :

ΔV mercury - ΔV glass : 5.76-0.384 = 5.376 cm³

7 0
3 years ago
The rate at which a certain Australian tree cricket chirps is 194/min at 28°C, but only 47.6/min at 5°C, From these data calcula
uysha [10]

Answer: The energy of activation for the chirping process is 283.911 kJ/mol

Explanation:

According to the Arrhenius equation,

K=A\times e^{\frac{-Ea}{RT}}

The expression used with catalyst and without catalyst is,

\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}]

where,

K_2 = rate of reaction at 28^0C = 194/min

K_1 = rate of reaction at  5^0C = 47.6 /min

Ea = activation energy

R = gas constant = 8.314 J/Kmol

tex]T_1[/tex] = initial temperature = 5^oC=273+5=278K

tex]T_1[/tex] = final temperature = 28^oC=273+28=301K

Now put all the given values in this formula, we get

\frac{194}{47.6}=\frac{E_a}{2.303\times 8.314}[\frac{1}{278}-\frac{1}{301}]

{E_a}=283911J/mol=283.911kJ/mol

Thus the energy of activation for the chirping process is 283.911 kJ/mol

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4 years ago
Which of the following statements correctly describes the asthenosphere???????????????<br> HELP ASAP
yulyashka [42]
That it is below the lithosphere and it is the remainder of the mantle or lower mantle
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Answer:

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8 0
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