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2 years ago

15

What volume and mass of steam at 100. °C and 1.00 atm would release the same amount of energy during condensation as 100. cm^3 o

f liquid watet would release during freezing?

1 answer:

viva [34]2 years ago

4 0

Check on google and make it a question Maybe itll appear

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A chemistry student needs 50.0 g of methyl acetate for an experiment. By consulting the CRC Handbook of Chemistry and Physics, t

slamgirl [31]

Explanation:

As density is defined as the mass of a substance divided by its volume.

Mathematically, Density = $\frac{mass}{volume}$

It is given that mass is 50 g and density is 0.934 $g/cm^{3}$ .

Hence, calculate the volume of methyl acetate as follows.

Density = $\frac{mass}{volume}$

0.934 $g/cm^{3}$ = $\frac{50 g}{volume}$

Volume = $53.53 cm^{3}$

or, = $53.53 ml$ (as 1 $cm^{3}$ = 1 mL)

Thus, we can conclude that the volume of methyl acetate the student should pour out is $53.53 ml$ .

3 0

2 years ago

Which is stronger - the attractive forces between water molecules and chromium and chloride ions, or the combined ionic bond str

Allisa [31]

Out of the two, the forces between water molecules and chromium and chloride ions is greater. This is proven by the fact that chromium chloride is slightly soluble in water, about 565 grams per liter.
In order for a substance to be soluble, the attraction of the ions to the water molecules must exceed the attraction between its own molecules and the water molecules.

3 0

2 years ago

The smallest particle in any element is a compound. *

zubka84 [21]

Answer:

false.

Explanation:

the smallest particle of a element is an atom

7 0

2 years ago

The atomic mass of helium-4 is 4.0026 amu. How many of each

umka21 [38]

Answer:

Protons: 2.

Electrons: 2.

Neutrons: 2.

Explanation:

Hello,

In this case, since an atom's atomic number is equal to the number of electrons, considering the electron configurations, taking into account that helium-4 is neither positively nor negatively charged, we can infer that the number of electrons equal the number of protons, which in this case are 2, due to the fact that is atomic number is 2.

Moreover, as helium-4's atomic number is 4 as a whole number, we compute the number of neutrons by using the shown below equation:

$Neutrons=mass\ number-atomic\ number\\\\Neutrons=4-2\\\\Neutrons=2$

Regards.

7 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