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zepelin [54]
3 years ago
4

What is the formula equation for the reaction between sulfuric acid and dissolved sodium hydroxide if all products and reactants

are in the aqueous or liquid phase?
Upper H upper C l (a q) plus upper N a upper O upper H (a q) right arrow upper N a upper C l (s) plus upper H subscript 2 upper O (l).
Upper H subscript 2 upper S upper O subscript 4 (a q) plus upper C a (upper O upper H) subscript 2 (a q q) right arrow upper C a upper S upper O subscript 4 (a q) plus 2 upper H subscript 2 upper O (l).
Upper H subscript 2 upper S upper O subscript 4 (a q) plus 2 upper N a upper O upper H (a q) right arrow upper N a subscript 2 upper A upper O subscript 4 (a q) plus 2 upper H subscript 2 upper O (l).
2 upper H subscript 3 upper P upper O subscript 4 (a q) plus 3 upper C a (upper O upper H) subscript 2 (a q) right arrow upper C a subscript 3 (upper PO upper O) subscript 4 (s) plus 6 upper H subscript 2 upper O (l).
Chemistry
2 answers:
Sonbull [250]3 years ago
7 0

Answer: The Answer is C.

Explanation: I just took the test.

Bezzdna [24]3 years ago
6 0

Answer:

H₂SO₄(aq) + NaOH(aq)   → Na₂SO₄(aq) +  H₂O(I)

Explanation:

Chemical equation:

H₂SO₄(aq) + NaOH(aq)   → Na₂SO₄(aq) +  H₂O(I)

Balanced chemical equation:

H₂SO₄(aq) + 2NaOH(aq)   → Na₂SO₄(aq) +  2H₂O(I)

It is an acid base reaction. Sulfuric acid and sodium hydroxide react to form the salt and water.

The reaction follow the law of conservation of mass because there are equal number of atoms on both side of equation.

According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Type of reaction:

It is double displacement reaction.

Double replacement:

It is the reaction in which two compound exchange their ions and form new compounds.

AB + CD → AC +BD

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Which combination of an element and an ion will react? View Available Hint(s) Which combination of an element and an ion will re
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<u>Answer:</u> The combination of element ad an ion that will react is Ni(s)\text{ and }Pt^{2+}(aq.)

<u>Explanation:</u>

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

X\rightarrow X^{n+}+ne^-

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

X^{n+}+ne^-\rightarrow X

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

For a reaction to be spontaneous, the standard electrode potential must be positive.

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

E^o_{cell}=E^o_{cathode}-E^o_{anode}       ......(1)

For the given options:

  • <u>Option 1:</u>  Sn(s)\text{ and }Mn^{2+}(aq.)

Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.

<u>Oxidation half reaction:</u>  Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V

<u>Reduction half reaction:</u>  Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V

Putting values in equation 1, we get:

E^o_{cell}=-1.18-(-0.14)=-1.04V

As, the standard potential is coming out to be negative, the given reaction will not take place.

  • <u>Option 2:</u>  Fe(s)\text{ and }Ca^{2+}(aq.)

Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.

<u>Oxidation half reaction:</u>  Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V

<u>Reduction half reaction:</u>  Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V

Putting values in equation 1, we get:

E^o_{cell}=-2.87-(-0.44)=-2.43V

As, the standard potential is coming out to be negative, the given reaction will not take place.

  • <u>Option 3:</u>  Ni(s)\text{ and }Pt^{2+}(aq.)

Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.

<u>Oxidation half reaction:</u>  Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V

<u>Reduction half reaction:</u>  Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V

Putting values in equation 1, we get:

E^o_{cell}=1.2-(-0.25)=1.45V

As, the standard potential is coming out to be positive, the given reaction will take place.

  • <u>Option 4:</u>  H_2(g)\text{ and }Na^{+}(aq.)

Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.

<u>Oxidation half reaction:</u>  H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V

<u>Reduction half reaction:</u>  Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V

Putting values in equation 1, we get:

E^o_{cell}=-0.27-(-0)=-0.27V

As, the standard potential is coming out to be negative, the given reaction will not take place.

Hence, the combination of element ad an ion that will react is Ni(s)\text{ and }Pt^{2+}(aq.)

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If 42.7 of 0.208 M hydrochloric acid are needed to completely neutralize a solution of calcium hydroxide, how many grams of calc
Montano1993 [528]

Answer:

0.329 g

Explanation:

In the context of this problem, we have a chemical reaction between hydrochloric acid and calcium hydroxide. HCl is the acid here and calcium hydroxide is the base. Hence, we have an acid-base reaction, also known as neutralization reaction.

In a neutralization reaction, water is produced as a product, as well as a salt that we obtain after we exchange the cations: calcium bonds to chloride and hydrogen bonds to hydroxide (the latter is the formation of water). This means we also produce calcium chloride as a product. The overall reaction represents this as:

Ca(OH)_2(aq)+2 HCl (aq)\rightarrow CaCl_2 (aq)+2 H_2O (l)

Firslt of all, we wish to find the number of moles of HCl present. Having molarity and volume, this is done by applying the molarity formula. It states that molarity is equal to the rate between moles and volume:

c_{HCl}=\frac{n_{HCl}}{V_{HCl}}

Rearranging for moles of HCl, n:

n_{HCl}=c_{HCl}V_{HCl}

Based on stoichiometry of the balanced chemical equation, notice that 1 mole of calcium hydroxide reacts with 2 moles of HCl, meaning:

n_{Ca(OH)_2}=\frac{1}{2} n_{HCl}=\frac{1}{2}c_{HCl}V_{HCl}

Now that we have the expression for moles, we may also express moles of calcium hydroxide as the ratio between its mass and molar mass:

n_{Ca(OH)_2}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}

Using the last two equations, we see that:

\frac{1}{2}c_{HCl}V_{HCl}=\frac{m_{Ca(OH)_2}}{M_{Ca(OH)_2}}\\\therefore m_{Ca(OH)_2}=\frac{1}{2}c_{HCl}V_{HCl}M_{Ca(OH)_2}

Substitute the given data, as well as the molar mass of calcium hydroxide:

m_{Ca(OH)_2}=\frac{1}{2}\cdot0.208 M\cdot0.0427 L\cdot74.093 g/mol=0.329 g

8 0
4 years ago
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