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Licemer1 [7]
3 years ago
9

Do Covalent bonds have weaker or stronger chemical bonds than Ionic bonds?

Chemistry
1 answer:
irina1246 [14]3 years ago
4 0

Ionic bonds are stronger than covalent bonds since they are of opposite charges. This makes them more attracted to each other, creating bonds that are harder to break. Covalent bonds don't have this and only occur to share electrons between atoms.

I hope this helps!

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When a liquid is heated, the temperature stops rising at the liquid’s ____________________.
Agata [3.3K]

Hi there,

the answer to the blank is: boiling point

When a liquid is heated, the temperature stops rising at the liquid's boiling point.

Hope this is correct :)

Have a great day

5 0
3 years ago
Read 2 more answers
Se tiene una solución acuosa 2M de carbonato de potasio. Expresar su concentración en %p/v y Normalidad.
zepelin [54]

Answer:

Normalidad = 4N

%p/V = 27.6%

Explanation:

La solución 2M de carbonato de potasio contiene 2moles de carbonato por litro de solución. La normalidad son los equivalente de carbonato de potasio (2eq/mol) por litro de solución:

2moles * (2eq/mol) = 4eq / 1L = 4N

El porcentaje peso volumen es el peso de carbonato en gramos dividido en el volumen en mL por 100:

%p/V:

Masa K2CO3 -Masa molar: 138.205g/mol-

2moles * (138.205g/mol) = 276g K2CO3

Volumen:

1L * (1000mL/1L) = 1000mL

%p/V:

276g K2CO3 / 1000mL * 100

<h3>%p/V = 27.6%</h3>
6 0
3 years ago
Examples of pure and impure substances
erastova [34]
Pure- table salt
Impure- vegetable oil
8 0
3 years ago
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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
3 years ago
A sample of gas has an initial pressure of 1.5 atm, an initial volume of 3.0 L, and an initial temperature of 293K. If the final
Nataliya [291]

Answer:

1.9 L

Explanation:

Step 1: Given data

  • Initial pressure (P₁): 1.5 atm
  • Initial volume (V₁): 3.0 L
  • Initial temperature (T₁): 293 K
  • Final pressure (P₂): 2.5 atm
  • Final volume (V₂): ?
  • Final temperature (T₂): 303 K

Step 2: Calculate the final volume of the gas

If we assume ideal behavior, we can calculate the final volume of the gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 1.5 atm × 3.0 L × 303 K / 293 K × 2.5 atm = 1.9 L

6 0
3 years ago
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