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Artemon [7]
3 years ago
8

AP CHEMISTRY -If any of these questions in the image (NET IONIC BALANCED EQUATIONS w/ states of matter) can be answered, especia

lly with how, would be amazing.

Chemistry
1 answer:
OLga [1]3 years ago
5 0

3) CH₃-COOH + NH₃ → CH₃-COO⁻NH₄⁺

4) 2 FeCl₃ + 3 Ag₂SO₃ → Fe₂(SO₃)₃ + 6 AgCl

5) 2 Al + 3 NiCl₂ → 2 AlCl₃ + 3 Ni

6) 4 LiCl + Pb(NO₂)₄ → 4 LiNO₂ + PbCl₄

7) 3 H₂SO₄ + 2 Al(OH)₃ → Al₂(SO₄)₃ + 6 H₂O

8) Cd(NO₃)₂ + Na₂S → CdS + 2 NaNO₃

9) Cr₂(SO₄)₃ + 3 (NH₄)₂CO₃ → Cr₂(CO₃)₃ + 3 (NH₄)₂SO₄

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What keeps gravitational force beteen two objects unchanged
hichkok12 [17]
Keeping the masses of the objects unchanged, if the distance between the objects is halved, then the magnitude of gravitational force between them will become. Hope this helps please mark brainliest :)
4 0
3 years ago
Use the given data at 500 K to calculate ΔG°for the reaction
Anton [14]

Answer : The  value of \Delta G^o for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)

First we have to calculate the enthalpy of reaction (\Delta H^o).

\Delta H^o=H_f_{product}-H_f_{reactant}

\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]

where,

\Delta H^o = enthalpy of reaction = ?

n = number of moles

\Delta H_f^0 = standard enthalpy of formation

Now put all the given values in this expression, we get:

\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]

\Delta H^o=-1035.6kJ=-1035600J

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction (\Delta S^o).

\Delta S^o=S_f_{product}-S_f_{reactant}

\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]

where,

\Delta S^o = entropy of reaction = ?

n = number of moles

\Delta S_f^0 = standard entropy of formation

Now put all the given values in this expression, we get:

\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]

\Delta S^o=-153J/K

Now we have to calculate the Gibbs free energy of reaction (\Delta G^o).

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

At room temperature, the temperature is 500 K.

\Delta G^o=(-1035600J)-(500K\times -153J/K)

\Delta G^o=-959100J=-959.1kJ

Therefore, the value of \Delta G^o for the reaction is -959.1 kJ

3 0
2 years ago
Find the normality of 0.321 g sodium carbonate in a 250 mL solution.
Radda [10]

Answer:

the normality of the given solution is 0.0755 N

Explanation:

The computation of the normality of the given solution is shown below:

Here we have to realize the two sodiums ions per carbonate ion i.e.

N = 0.321g Na_2CO_3 × (1mol ÷ 105.99g)×(2eq ÷ 1mol)

= 0.1886eq ÷ 0.2500L

= 0.0755 N

Hence, the normality of the given solution is 0.0755 N

5 0
2 years ago
Gregor Mendel identified seven distinguishable individual __________________________Each one he discovered had two different for
Eddi Din [679]

Answer:

objects

Explanation:

6 0
3 years ago
For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of
torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For Al_2O_3

Mass of Al_2O_3  = 4.07 g

Molar mass of Al_2O_3  = 101.96 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{4.07\ g}{101.96\ g/mol}

Moles\ of\ Al_2O_3= 0.0399\ mol

According to the reaction:

Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O

1 mole of Al_2O_3  on reaction produces 1 mole of Al_2(SO_4)_3

So,  

0.0399 mole of Al_2O_3  on reaction produces 0.0399 mole of Al_2(SO_4)_3

Moles of Al_2(SO_4)_3  obtained = 0.0399 mole

Molar mass of Al_2(SO_4)_3 = 342.2 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0399= \frac{Mass}{342.2\ g/mol}

Mass= 13.7\ g

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

\%\ yield =\frac{10.4}{13.7}\times 100

<u>% yield =76 %</u>

6 0
2 years ago
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