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3 years ago

What is the difference between active transport and passive transport? Which one requires energy?

Chemistry

1 answer:

Marta_Voda [28]3 years ago

4 0

Active is currently and passive is no longer active. Active requires energy

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Where are non-metals located on the periodic table on the right of the periodic table after the staircase where would you find t

djyliett [7]

Did you mean why? not where?

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3 years ago

Carbonate oxidation number

Svetllana [295]

Answer:

-2

Explanation:

carbonate oxifation number is -2

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3 years ago

If a piece of cadmium with a mass of 37.60 g and a temperature of 100.0 oC is dropped into 25.00 cc of water at 23.0 oC, what wi

zlopas [31]

Answer:

$T_{eq}=28.9\°C$

Explanation:

Hello!

In this case, since it is observed that hot cadmium is placed in cold water, we can infer that the heat released due to the cooling of cadmium is gained by the water and therefore we can write:

$Q_{Cd}+Q_{W}=0$

Thus, we insert mass, specific heat and temperatures to obtain:

$m_{Cd}C_{Cd}(T_{eq}-T_{Cd})+m_{W}C_{W}(T_{eq}-T_{W})=0$

In such a way, since the specific heat of cadmium and water are respectively 0.232 and 4.184 J/(g °C), we can solve for the equilibrium temperature (the final one) as shown below:

$T_{eq}=\frac{m_{Cd}C_{Cd}T_{Cd}+m_{W}C_{W}T_{W}}{m_{Cd}C_{Cd}+m_{W}C_{W}}$

Now, we plug in to obtain:

$T_{eq}=\frac{37.60g*0.232\frac{J}{g\°C}*100.00\°C+25.00g*4.184\frac{J}{g\°C}*23.0\°C}{37.60g*0.232\frac{J}{g\°C}+25.00g*4.184\frac{J}{g\°C}}\\\\T_{eq}=28.9\°C$

NOTE: since the density of water is 1g/cc, we infer that 25.00 cc equals 25.00 g.

Best regards!

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3 years ago

A student was asked to find the molar mass of calcium nitrate, Ca(NO3)2. she added the molar masses of calcium, nitrogen, and ox

dlinn [17]

Answer: 116g/mole

Explanation:

She didn't get the answer because she didn't add the them well , due to the bracket present.

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3 years ago

Predict the effect of adding a non competitive inhibitor to the reaction mixture on the rate of reaction at a high substrate con

Darya [45]

Answer:

A noncompetitive inhibitor can only bind to an enzyme with or without a substrate at several places at a particular point in time

Explanation:

this is because It changes the conformation of an enzyme as well as its active site, which makes the substrate unable to bind to the enzyme effectively so that the efficiency of the enzyme decreases. A noncompetitive inhibitor binds to the enzyme away from the active site, altering/distorting the shape of the enzyme so that even if the substrate can bind, the active site functions less effectively and most of the time also the inhibitor is reversible

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3 years ago