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Lisa [10]
3 years ago
7

In an experiment, a student places a small piece of pure Mg(s) into a beaker containing 250.mL of 6.44MHCl(aq) . A reaction occu

rs, as represented by the equation above. (a) Write the balanced net ionic equation for the reaction between Mg(s) and HCl(aq) .
Chemistry
1 answer:
solong [7]3 years ago
8 0

Answer:

Mg(s) + 2H^+(aq) ----> Mg^2+(aq) + H2(g)

Explanation:

We must first write the molecular equation, then obtain the complete ionic equation and subsequently get the net ionic equation as shown below;

Molecular reaction equation;

Mg(s) + 2HCl(aq) ------> MgCl2(aq) + H2(g)

Complete ionic equation;

Mg(s) + 2H^+(aq) + 2Cl^-(aq) ----> Mg^2+(aq) + 2Cl^-(aq) + H2(g)

Net ionic equation;

Mg(s) + 2H^+(aq) ----> Mg^2+(aq) + H2(g)

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Among the elements listed, which would show the largest increase between the second and third ionization energies and why?
Yuri [45]
Ionization energy increases from left to right in the row and from bottom to top in a column. Also as we get closer to the nucleus it would be harder to take electrons out. B  (atomic #5) has 2 layers of electron 2 and 3 atom in each layer. P has 15 so it would be 2,8 and 5 respectively. Ca is 20 so 2,8,8,2 and Zn is 30 and it would be 2,18,8,2.
For energy between second and third ionization we are looking at taking out the 3rd electron. B already has 3 electron in the first layer so its easy to take them all. P has 5 in the last layer so again easy. But when we look at Ca and Zn after the 2nd electron (in the last layer) we should change the layer go one layer inside. So this needs more energy. To pick between Zn and Ca (they are in the same row)  I mentioned earlier that in one row as we go to the right ionization energy increases so the answer is Zn. 
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3 years ago
Captures sunlight to make food for the cell through photosynthesis.
melisa1 [442]

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Chloroplasts

Explanation:

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4 0
2 years ago
Calculate the equilibrium constant for the following reaction: Co2+ (aq) + Zn(s> CO (s) + Zn2+ (aq)
Simora [160]

<u>Answer:</u> The K_{eq} of the reaction is 1.73\times 10^{16}

<u>Explanation:</u>

For the given half reactions:

Oxidation half reaction: Zn(s)\rightarrow Zn^{2+}+2e^-;E^o_{Zn^{2+}/Zn}=-0.76V

Reduction half reaction: Co^{2+}+2e^-\rightarrow Co(s);E^o_{Co^{2+}/Co}=-0.28V

Net reaction: Zn(s)+Co^{2+}\rightarrow Zn^{2+}+Co(s)

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

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

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

Putting values in above equation, we get:

E^o_{cell}=-0.28-(-0.76)=0.48V

To calculate equilibrium constant, we use the relation between Gibbs free energy, which is:

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

and,

\Delta G^o=-RT\ln K_{eq}

Equating these two equations, we get:

nfE^o_{cell}=RT\ln K_{eq}

where,

n = number of electrons transferred = 2

F = Faraday's constant = 96500 C

E^o_{cell} = standard electrode potential of the cell = 0.48 V

R = Gas constant = 8.314 J/K.mol

T = temperature of the reaction = 25^oC=[273+25]=298K

K_{eq} = equilibrium constant of the reaction = ?

Putting values in above equation, we get:

2\times 96500\times 0.48=8.314\times 298\times \ln K_{eq}\\\\K_{eq}=1.73\times 10^{16}

Hence, the K_{eq} of the reaction is 1.73\times 10^{16}

8 0
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Answer:

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Explanation:

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If the polymer chains are close together such that the resultant polymer is crystalline with chains packed closely, we have high density polyethylene.

On the other hand if the chains are not close together and there are spaces left between chains when the polymer is formed, then we have low density polyethylene.

Hence, low density polyethylene gets its name from the spaces left between chains when the polymer is formed.

6 0
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