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kiruha [24]
3 years ago
12

Please help bestossss

Chemistry
1 answer:
Yuki888 [10]3 years ago
3 0

Answer:

it's option A.

cell, tissue, organ, system

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I need helppp plssss
Ganezh [65]
Li2O + H2O → 2LiOH
This is the answer
3rd blank write 2
3 0
3 years ago
A clam farmer has been keeping records of the water temperature and the number of clams developing from fertilized eggs. The dat
adoni [48]

Answer:

The dependent variable is the number of clams developing from fertilized eggs.

The independent variable is the water temperature

The optimum temperature for clam development is 30 degrees centigrade.

Explanation:

The graph of the number of clams developing from fertilized eggs and water temperature is attached to this answer.

The independent variable is being manipulated in an experiment. As it changes, it produces a corresponding change in the dependent variable.

Here, the water temperature is the independent variable. As it changes, the number of clams developing from fertilized eggs (dependent variable) also changes alongside.

The optimum temperature is the temperature at which the greatest number of clams developing from fertilized eggs is produced. We can see from the graph that this temperature is 30 degrees centigrade.

6 0
3 years ago
Potential (P) energy or kinetic (K) energy
JulsSmile [24]
Potential energy, kinetic energy would be if they were already running
8 0
3 years ago
Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca
lina2011 [118]

<u>Answer:</u> The above reaction is non-spontaneous.

<u>Explanation:</u>

For the given chemical reaction:

Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.23-0.77=-1.0V

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

3 0
3 years ago
Вычислить равновесные концентрации частиц в растворе, содержащем 0,01 моль/л Cu(NO3)2 и 1,0 моль/л NH3.
Wewaii [24]

Answer:

i dOnT SpEaK uR lAnGuAgE

Explanation:

ReEEEeEeEeEeEeEeEeEeeEEEeeeeEEEEeeeeeEEeEEeeeeeEEEEEeeeEee

3 0
3 years ago
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