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Effectus [21]
3 years ago
9

Vinegar produces carbon dioxide gas when combined with baking soda. Physical or chemical property?

Chemistry
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

it's a chemical property

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<span>the table say that at 20 degree celcius 88.0g of NANO3 will remain dissolved in 100 gm of H2O so at 20 degree celcius 80.0g of H20 will dissolve (88.0g)x(80g/100g)=70.4g of NANO3 so at 20 degree celcius 86.3g-70.4g= 15.9 gram of NANO3 will come out of solution.</span>
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How can a short-term environmental change, such as a drought or a flood, affect organisms?
viktelen [127]

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it can affect things by drastically chagning the way that organisms opareate such as the eco systems, the health of the land the flood or drought is on and etc.

Explanation:

hope this helps!

5 0
2 years ago
What is the orbital or electronic geometry of a molecule with 0 nonbonding electron pairs and 2 bonding electron pairs?
disa [49]
Linear is the answer
3 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
3 years ago
As the mass of a sample increases, the number of moles present in the sample
vivado [14]

Answer:

increases

Explanation:

4 0
3 years ago
Read 2 more answers
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