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snow_tiger [21]
3 years ago
7

How do the properties of table salt compare with those of sodium and chlorine?

Chemistry
1 answer:
Mashutka [201]3 years ago
5 0

Well practically sodium and chlorine are both poisonous so mixing two poisons tastes wonderful on food.

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Perform the following calculation and report the answer with the correct number of significant figures.
Paul [167]

Answer:

81.72 +0.73=82.45

Explanation:

  • you need to add it

1

81.72

+0.73

= 82.45

8 0
2 years ago
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What is the mass of a liquid that has a volume of 50.0ml and a density of 2.40g/ml?
Iteru [2.4K]

Answer:

120g Using Density Equation.

Explanation:

Density =mass/volume

We need to the solve for the mass.

Mass=Density * Volume

Mass= 2.40 g/ml * 50.0 ml = 120g

5 0
3 years ago
What occurs as energy is transferred through the convective zone of the Sun? Check all that apply.
vlada-n [284]
-Photons are absorbed by hot gas atoms

-Energy is transferred through large-scale movement of material

-Energy is released into the photosphere
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3 years ago
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True or false? You can follow the progress of a reaction that produces a gas using a sensitive mass
sasho [114]

Answer:

true

Explanation:

3 0
2 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
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