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4 years ago

How are nymphs different from adult insects?

Chemistry

2 answers:

umka2103 [35]4 years ago

4 0

Answer:

Explanation:

the nymph looks like a smaller version of the adult insect, & it also has no wings & molts.

Ede4ka [16]4 years ago

3 0

Answer:

tg n

Explanation:

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The diagram below represents a plan cell Letter X represents a structure in the cell

weeeeeb [17]

the cell structure represented by x is the nucleus

4 0

3 years ago

The estimated heat of vaporization of diethyl ether using the Chen's rule is A. 29.7 KJ/mol B. 33.5 KJ/mol C. 26.4 KJ/mol D. 36.

Brums [2.3K]

Answer:

C. 26.4 kJ/mol

Explanation:

The Chen's rule for the calculation of heat of vaporization is shown below:

$\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]$

Where,

$\Delta H_v$ is the Heat of vaoprization (J/mol)

$T_b$ is the normal boiling point of the gas (K)

$T_c$ is the Critical temperature of the gas (K)

$P_c$ is the Critical pressure of the gas (bar)

R is the gas constant (8.314 J/Kmol)

For diethyl ether:

$T_b=307.4\ K$

$T_c=466.7\ K$

$P_c=36.4\ bar$

Applying the above equation to find heat of vaporization as:

$\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]$

$\Delta H_v=26400 J/mol$

The conversion of J into kJ is shown below:

1 J = 10⁻³ kJ

Thus,

$\Delta H_v=26.4 kJ/mol$

Option C is correct

6 0

4 years ago

What type of chemical reaction is oil spills?

labwork [276]

Decomposition, because the oil eventually breaks down bc of bacterias and evaporate into the ocean!

8 0

2 years ago

A 48.0g sample of quartz, which has a specific heat capacity of 0.730·J·g−1°C−1, is dropped into an insulated container containi

Butoxors [25]

Answer:

The equilibrium temperature of the water is 26.7 °C

Explanation:

Step 1: Data given

Mass of the sample quartz = 48.0 grams

Specific heat capacity of the sample = 0.730 J/g°C

Initial temperature of the sample = 88.6°C

Mass of the water = 300.0 grams

Initial temperature = 25.0°C

Specific heat capacity of water = 4.184 J/g°C

Step 2: Calculate final temperature

Qlost = -Qgained

Qquartz = - Qwater

Q =m*c*ΔT

Q = m(quartz)*c(quartz)*ΔT(quartz) = -m(water) * c(water) * ΔT(water)

⇒ mass of the quartz = 48.0 grams

⇒ c(quartz) = the specific heat capacity of quartz = 0.730 J/g°C

⇒ ΔT(quartz) = The change of temperature of the sample = T2 -88.6 °C

⇒ mass of water = 300.0 grams

⇒c(water) = the specific heat capacity of water = 4.184 J/g°C

⇒ ΔT= (water) = the change in temperature of water = T2 - 25.0°C

48.0 * 0.730 * (T2-88.6) -300.0 * 4.184 *(T2 - 25.0)

35.04(T2-88.6) = -1255.2 (T2-25)

35.04T2 -3104.544 = -1255.2T2 + 31380

1290.24T2 = 34484.544

T2 = 26.7 °C

The equilibrium temperature of the water is 26.7 °C

8 0

3 years ago

You have a mass of 54 kg and are sitting on a stool. The normal force the stool applies on you is ...

Karo-lina-s [1.5K]

Answer:

529.2 N

Explanation:

As we have studied the first law of motion, which states that every action has some reaction, equal in magnitude but having an opposite direction.

The force that is acting on the student will be due to gravitational force, that is equal to his weight.

F=mg: 54kg x 9.8m/s^2 =529.2 N

So the weight of student is exerting downwards towards the stool and land. The stool will also exert a force on the student that will be equal in magnitude but opposite in direction, then it will be 529.2 N.

This is because the student is sitting in a constant state and all the weight is exerted on the stool.

Note: This answer is very generic supposing that all the weight of the student is on stool. But, if we suppose that student's legs are on floor so it means the force of gravity acting on the stool has become less because student's mass on stool is less. So the answer would be a force somehow less than 529.2 N. However, since the question asked normal force, it would be weight of student in general terms.

Hope it helps!

7 0

3 years ago