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2 years ago

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Why does salad become soggy and wilted when the dressing has been on it for a while? Explain why, in terms of osmosis. PLS HELP

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1 answer:

musickatia [10]2 years ago

5 0

This means the water from the leaves goes into the dressing, causing the leaves to appear wilted.

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How to solve this: a small dog has consumed a dangerous amount of apap. the vet has ordered: rx acetylcysteine 150mg/kg/stat. yo

Fantom [35]

Given:
Stock dose/concentration of 20% Acetylcysteine (200 mg/mL)
150 mg/kg dose of Acetylcysteine
Weight of the dog is 13.2 lb

First we must convert 13.2 lb to kg:
13.2 lb/(2.2kg/lb) = 6 kg

Then we must calculate the dose:
(150 mg/kg)(6kg) = 900 mg

Lastly, we must calculate the dose in liquid form to be administered:
(900 mg)/(200 mg/mL) = 4.5 mL

Therefore, 4.5 mL of 20% Acetylcysteine should be given.

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3 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:

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

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