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lorasvet [3.4K]
2 years ago
10

Write ionic equations for any three of the following:

Chemistry
1 answer:
worty [1.4K]2 years ago
5 0

1.

The ionic equation is:

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

HNO₃(aq) + NaOH(aq) → NaNO₃(aq) + H2O(l)

The required dissociations are

HNO₃(aq) → H⁺(aq) + NO₃⁻(aq)

NaOH(aq) → Na⁺(aq) + OH⁻(aq)

NaNO₃(aq) → Na⁺(aq) + NO₃⁻(aq)

So, the ionic equation is

H⁺(aq) + NO₃⁻(aq) + Na⁺(aq) + OH⁻(aq) → Na⁺(aq) + NO₃⁻(aq) + H₂O(l)

2.

The ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

HCl(aq) + KOH(aq) → KCl(aq) + H2O(l)

The required dissociations are

HCl(aq) → H⁺(aq) + Cl⁻(aq)

KOH(aq) → K⁺(aq) + OH⁻(aq)

KCl(aq) → K⁺(aq) + Cl⁻(aq)

So, the ionic equation is

H⁺(aq) + Cl⁻(aq) + K⁺(aq) + OH⁻(aq) → K⁺(aq) + Cl⁻(aq) + H₂O(l)

3.

The ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

H₂SO₄(aq) + Mg(OH)₂(aq) → MgSO₄(aq) + 2H2O(l)

The required dissociations are

H₂SO₄(aq) → 2H⁺(aq) + SO₄²⁻(aq)

Mg(OH)₂(aq) → Mg²⁺(aq) + 2OH⁻(aq)

MgSO₄(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

So, the ionic equation is

2H⁺(aq) + SO₄²⁻(aq) + Mg²⁺(aq) + 2OH⁻(aq) → Mg²⁺(aq) + SO₄²⁻(aq)

+ 2H₂O(l)

Learn more about ionic equations here:

brainly.com/question/11628165

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Answer:

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It will be observed after the second development is complete that the new Rf values will be 0.50 and 0.52 respectively. It will just be as though the second chromatogram picked up from where the first chromatogram stopped.

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Answer:

a. ΔH^0_{rxn} = -108.0\frac{kJ}{mol}

b. 320.76° C

Explanation:

a.)

we can solve this type of question (i.e calculate ΔH^0_{rxn} , for the gas-phase reaction )  using the Hess's Law.

ΔH^0_{rxn} =  E_{product} deltaH^0_{t}-E_{reactant} deltaH^0_{t}

Given from the question, the table below shows the corresponding  ΔH^0_{t}(kJ/mol) for each compound.

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Liquid EO                       -77.4

CH_4_(g_)                            -74.9                

CO_(g_)                              -110.5

If we incorporate our data into the above previous equation; we have:

ΔH^0_{rxn} = (-110.5 kJ/mol + (-74.9 kJ/mol) ) - (-77.4 kJ/mol)

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b.)

We are to find the final temperature if the average specific heat capacity of the products is 2.5 J/g°C

Given that:

the specific heat capacity (c) = 2.5 J/g°C

T_{initial} = 93.0°C   &

the  enthalpy of vaporization  (ΔH^0_{vap}) = 569.4 J/g

If, we recall; we will remember that; Specific Heat Capacity is the amount of heat needed to raise the temperature of one gram of a substance by one kelvin.

∴ the specific heat capacity (c) is given as =  \frac{Heat(q)}{mass*changeintemperature(T_{initial}-T_{final})}

Let's not forget as well, that  ΔH^0_{vap} = \frac{q}{mass}

If we substitute  ΔH^0_{vap}  for  \frac{q}{mass} in the above equation, we have;

specific heat capacity (c) = \frac{deltaH^0_{vap}}{T_{final}-T_{initial}}

Making (T_{final}- T_{initial}) the subject of the formula; we have:

T_{final}- T_{initial}  = \frac{delat H^0_{vap}}{specificheat capacity}

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T_{final}=\frac{569.4J/g}{2.5J/g^0C}+93.0^0C

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          = 320.76°C

∴ we can thereby conclude that the final temperature = 320.76°C                

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