The U.S. Integrated Ocean Observing System (IOOS) was created to

Chemistry

1 answer:

belka [17]4 years ago

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Answer:

A I believe this is the answer I do not know look at the picture I attached

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Glaxo Smith Kline (GSK) in Research Triangle Park has hired a recent biology graduate to explore a drug that would inactive memb

Rainbow [258]

Answer:

Glucose would not be able to get into the cell treated with this chemical

Potassium mwill not be able to move or transport may be affected

Explanation:

As all membrane transport proteins are inactivated glucose is abig molecule it cannot pass without transporter protein.

Potassium transport is through sodium potassium pump and leak channels. As all transport protein are affected so it should not be able to move but if drug does not affect them then they will be unaffected.

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4 years ago

Compare the properties of protons, neutrons and electrons

sergejj [24]

Answer:

Electrons are a type of subatomic particle with a negative charge. ... Protons and neutrons have approximately the same mass, but they are both much more massive than electrons (approximately 2,000 times as massive as an electron). The positive charge on a proton is equal in magnitude to the negative charge on an electron

Explanation:

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4 years ago

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A galvanic cell consists of one half-cell that contains Ag(s) and Ag+(aq), and one half-cell that contains Cu(s) and Cu2+(aq). W

Agata [3.3K]

Answer : The 'Ag' is produced at the cathode electrode and 'Cu' is produced at anode electrode under standard conditions.

Explanation :

Galvanic cell : It is defined as a device which is used for the conversion of the chemical energy produces in a redox reaction into the electrical energy. It is also known as the voltaic cell or electrochemical cell.

In the galvanic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

We are taking the value of standard reduction potential form the standard table.

$E^0_{[Ag^{+}/Ag]}=+0.80V$

$E^0_{[Cu^{2+}/Cu]}=+0.34V$

In this cell, the component that has lower standard reduction potential gets oxidized and that is added to the anode electrode. The second forms the cathode electrode.

The balanced two-half reactions will be,

Oxidation half reaction (Anode) : $Cu(s)\rightarrow Cu^{2+}(aq)+2e^-$