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zepelin [54]
2 years ago
14

A 55.0-mg sample of al(oh)3 is reacted with 0.200m hcl. how many milliters of the acid are needed to neutralize the al(oh)3?

Chemistry
1 answer:
chubhunter [2.5K]2 years ago
8 0
Al(OH)₃ + 3HCl = AlCl₃ + 3H₂O

C=0.200 mmol/mL
M{Al(OH)₃}=78.0 mg/mmol
m{Al(OH)₃}=55.0 mg

n{Al(OH)₃}=m{Al(OH)₃}/M{Al(OH)₃}

3n{Al(OH)₃}=n(HCl)=CV

3m{Al(OH)₃}/M{Al(OH)₃}=CV

V=3m{Al(OH)₃}/[CM{Al(OH)₃}]

V=3*55.0/[0.200*78.0]=10.6 mL
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OLga [1]

Answer is: A. Chemical energy to electromagnetic energy and thermal energy.  

Balanced chemical reaction: 2Mg(s) + O₂(g) → 2MgO(s) + energy.

This is chemical change (chemical reaction), because new substance (magnesium oxide MgO) is formed, the atoms are rearranged and the reaction is followed by an energy change (exothermic reaction because energy is released).

Chemical changes (chemical synthesis) is when a substance combines with another (in this example magnesium and oxygen) to form a new substance.


5 0
2 years ago
Read 2 more answers
A sample of an unknown compound was decomposed and found to be composed of 1.36 mol oxygen, 4.10 mol hydrogen, and 2.05 mol carb
Flura [38]

Answer:

C3H6O2

Explanation:

To find the empirical formula of the compound, we divide the amount in moles of each of the elements by the amount in mole of the element with the smallest number of mole. In this question, the element with the smallest number of moles is oxygen with 1.36 mole. Hence, we divide the number of moles of each element by this.

H = 4.10/1.36 = 3

O = 1.36/1.36 = 1

C = 2.05/1.36 = 1.5

We then multiply through by 2 to yield the compound with the empirical formula C3H6O2

7 0
2 years ago
Complete ionic equation K2CO3(aq)+2CuF(aq) → Cu2CO3(s)+2KF(aq) Examine each of the chemical species involved to determine the io
Fudgin [204]

Answer:

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

Explanation:

K2CO3(aq) + 2CuF(aq) → Cu2CO3(s) + 2KF(aq)

The complete ionic equation for the above equation can be written as follow:

In solution, K2CO3 and CuF will dissociate as follow:

K2CO3(aq) —› 2K+(aq) + CO3²¯(aq)

CuF(aq) —› Ca^2+(aq) + 2F¯(aq)

Thus, we can write the complete ionic equation for the reaction as shown below:

K2CO3(aq) + 2CuF(aq) —›

2K+(aq) + CO3²¯(aq) + Ca^2+(aq) + 2F¯(aq) —› Cu2CO3(s) + 2K+(aq) + 2F¯(aq)

8 0
2 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
2 years ago
Which of the following describes the electron sharing between hydrogen and fluorine?
EastWind [94]
Hello
We know both the elements involved in bonding are non-metals and the primary type of bonds involved in non-metals are covalent bonds. Covalent bonds are formed when two atoms share one or more electrons; thus we know that whatever the number of electrons shared, it has to be equal for both. We can eliminate choices A and B.
Next, we understand that it is easier for one atom to be shared among the two, rather than the 7. First, because Hydrogen needs only 1 electron to be stable and would require energy to compensate the remaining 6. Second, electrons are attracted towards the nucleus so it is inefficient to try and share 7 electrons when 1 provides the same amount of stability to Fluorine.
Therefore, the answer is C.
4 0
2 years ago
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