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3 years ago

8

A synthetically-produced narcotic medication, drug, or agent similar to the opiate morphine, but not derived from opium; used to

relieve pain

1 answer:

Otrada [13]3 years ago

4 0

A synthetically-produced narcotic medication, drug, or agent similar to the opiate morphine, but not derived from opium; used to relieve pain
Yes

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1. What do the ice ages tell us about how Earth's<br> climate has changed in the past?

Levart [38]

Answer:

During ice ages, the most characteristic change to the planet has been the formation and spread large ice sheets and glaciers across much the Northern Hemisphere.

3 0

2 years ago

What happens to population growth in a logistic growth pattern as it reaches carrying capacity?

Leno4ka [110]

I believe the answer would be C. slows down, hope this helps:)

4 0

2 years ago

A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %

kondor19780726 [428]

This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.

In this case, it is recommended to write the enthalpy for each substance as follows:

$H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)$

Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and $y_{C-6}$ and $y_{N_2}$ are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

$H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol$

It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:

$Q=-40036+(-672.21)=-40708.21J$

Finally we convert this result to kJ:

$Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ$

Learn more:

6 0

1 year ago

A graduated cylinder contains 18.0ml of water. what is the new water level after 35.6g of silver metal with a density of 10.5 g/

AlekseyPX

Answer:

The answer to your question is

1.- Volume = 3.4 ml

2.- Volume = 0.61 ml

3.- Mass = 2872.8 pounds

Explanation:

Problem 1

Volume = 18 ml

mass = 35.6 g

density = 10.5 g/ml

Process

1.- Calculate the volume of silver

Formula

$density = \frac{mass}{volume}$

solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{35.6}{10.5}$

<u>volume = 3.4 ml</u>

2.- Problem 2

Total volume = ?

Volume = 18 + 3.4

Volume = 21.4 ml

Data

mass = 8.3 g

density = 13.6 g(ml

volume = ?

Formula

$density = \frac{mass}{volume}$

Solve for volume

$volume = \frac{mass}{density}$

Substitution

$volume = \frac{8.3}{13.6}$

Result

<u>volume = 0.61 ml</u>

3.- Problem 3

Data

volume = 345 gal

density = 1 g/ml

mass = ?

Formula

$density = \frac{mass}{volume}$

Solve for mass

mass = density x volume

Covert gal to ml

1 gal --------------- 3785 ml

345 gal ------------- x

x = (345 x 3785) / 1

x = 1305825 ml

Substitution

mass = 1 x 1305825

mass = 1305825 g

Convert g to pounds

1 g ------------------- 0.0022 pounds

1305825 g ---------------- x

x = (1305825 x 0.0022)

<u> x = 2872.8 pounds</u>

5 0

2 years ago

8. Sulfur has a first ionization energy of 1000 kJ/mol. Photons of what frequency are required to ionize one mole of Sulfur?

Lynna [10]

Answer:

the frequency of photons $v = 1.509\times10^{39}Hz$

Explanation:

Given: first ionization energy of 1000 kJ/mol.

No. of moles of sulfur = 1 mole

$\Delta E_1 = 1000KJ/mol$

We know that plank's constant

$h = 6.626\times10^{-34} Js$

Let the frequency of photons be ν

Also we know that ΔE = hν

this implies ν = ΔE/h

$= \frac{10^6J}{6.626\times10^{-34} Js}$

$v = 1.509\times10^{39}Hz$

Hence, the frequency of photons $v = 1.509\times10^{39}Hz$

6 0

2 years ago