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Harrizon [31]
3 years ago
6

How many grams of water can be formed when 80 grams of sodium hydroxide (NaOH) reacts with an excess of sulfuric acid, H2SO4, in

solution?
2NaOH + H2SO4 --> Na2SO4 + 2H2O

_____ grams water
Chemistry
1 answer:
podryga [215]3 years ago
5 0

Answer:

36g of H2O.

Explanation:

We'll begin by writing the balanced equation for the reaction:

2NaOH + H2SO4 —> Na2SO4 + 2H2O

Next, we shall determine the mass of NaOH that reacted and the mass of H2O produced from the balanced equation. This is illustrated below:

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass of NaOH from the balanced equation = 2 x 40 = 80g

Molar mass of H2O = (2x1) + 16 = 18g/mol

Mass of H2O from the balanced equation = 2 x 18 = 36g.

From the balanced equation above, we can see evidently that:

80g of NaOH reacted to produce 36g of H2O.

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Consider the incomplete structure. Add formal charges as necessary to the structure. All unshared valence electrons are shown.
Zina [86]

The net charge on the structure as shown in the question is -1.

<h3>What is a charged specie?</h3>

We say that a specie is charged if the specie has an excess of positive or negative charge. An excess of the positive charge means that the substance is positively charged while an excess if the negative charge simply means that the object is negative charged.

When we have a chemical structure as we have seen, it is possible that the structure as we have it could have a net charge. The net charge that the structure has can be deduced by looking at the formal charges that are carried by all the atoms that we have in the system.

The charge as we can see that is on the central atom of the molecule is the -1 charge hence this is the charge that is carried overall by the molecule.

If we then look at the structure as we can see, we can see that there is a charge of negative one that is attached to the atom of chlorine as is clearly visible and obvious from the image attached to the question.

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5 0
1 year ago
For single bonds between similar types of atoms, how does the strength of the bond relate to the sizes of the atoms? Explain.
uysha [10]

When comparing single bonds between atoms of comparable types, the stronger the bond is, the bigger the atom, the weaker it is.

The length of the X-H bond lengthens while the strength of the bond shortens with increasing halogen size (F-H strongest, I-H weakest). When comparing single bonds between atoms of similar sorts, the larger the atom, the weaker the bond. It can be explained by the fact that less energy is required to break the bond the bigger the atom's atomic size. The force of attraction from the nucleus to the outermost orbit will be less for iodine since it has a larger atom than the other elements in the group.

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7 0
1 year ago
The enthalpy of vaporization of liquid water at 100°C is 2257 kJ/kg. Determine the enthalpy for apordato of iuod eeat capacity o
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Explanation:

The given data is as follows.

         T_{1} = 100^{o}C,       T_{2} = 10^{o}C

       \Delta H_{vap1} = 2257 kJ/kg,     \Delta H_{vap2} = ?

For water, C_{p} = 4.184 kJ/kg ^{o}C

Formula to calculate heat of vaporization is as follows.

  \Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

Hence, putting the values into the above formula as follows.

\Delta H_{vap1} - \Delta H_{vap2} = C_{p}(T_{1} - T_{2})

2257 kJ/kg - \Delta H_{vap2} = 4.184 kJ/kg ^{o}C (100 - 10)^{o}C

            \Delta H_{vap2} = 2257 kJ/kg - 376.56 kJ/kg

                                       = 1880.44 kJ/kg

Thus, we can conclude that enthalpy of liquid water at 10^{o}C is 1880.44 kJ/kg.

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HELP SO MUCH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!! Read the following chemical equations.
Pavlova-9 [17]

Answer:

I is oxidized in reaction 1 and Cl2 is reduced in reaction 2.

Explanation:

The oxidation reduction reactions are called redox reaction. These reactions are take place by gaining or losing the electrons and oxidation state of elements are changed.

Oxidation:

Oxidation involve the removal of electrons and oxidation state of atom of an element is increased.

Reduction:

Reduction involve the gain of electron and oxidation number is decreased.

Consider the following reactions.

1 =    2KI  + H₂O₂    →       2KOH + 1₂

In this reaction the oxidation state of iodine is -1 on left hand side and 0 on right hand side which means it lost the electron and gets oxidized.

2=      Cl₂ + H₂       →       2HCl

In this reaction the oxidation state of chlorine on left hand side is 0 while on right hand side its -1 thus it gets reduced.

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