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Stolb23 [73]
3 years ago
14

Which of the following represents the least number of molecules?

Chemistry
1 answer:
lozanna [386]3 years ago
6 0

Answer:

A ; 20g of water has the least number of molecules

Explanation:

Here, we want to know which of the options represent the least number of molecules;

To calculate this, we are going to first calculate the number of moles using the formula below;

Number of moles = mass/molar mass

Then from the number of moles, we can calculate the number of molecules by multiplying the number of moles by 6.02 * 10^23 molecules

Let’s proceed;

a. That will be 20/18.02 = 1.1099 moles

1.1099 * 6.02 * 10^23 = 6.68 * 10^23 molecules

b. That will be 77/16.06 = 4.8 moles

4.8 * 6.02 * 10^23 = 2.89 * 10*24 molecules

c. That will be 68/42.09 = 1.62 moles

1.62 * 6.02 * 10^23 = 9.73 * 10^23 molecules

d. That will be 100/44.02 = 2.27 moles

2.27 * 6.02 * 10^23 = 1.37 * 10^24 molecules

e. That will be = 84/20.01 = 4.2 moles

4.2 * 6.02 * 10^23 = 2.53 * 10^24 molecules

From all the values obtained, the lowest is 20g of water

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Translation:

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Hydrogen gas and fluorine gas will react to form hydrogen fluoride gas. What is the standard free energy change for this reactio
Marta_Voda [28]

Answer:

\Delta G=-541.4kJ/mol

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

H_2+F_2\rightarrow 2HF

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

\Delta G=2\Delta G_f^{HF}-(\Delta G_f^{H_2}+\Delta G_f^{F_2})\\\\\Delta G=2*-270.70kJ/mol-(0kJ/mol+0kJ/mol)\\\\\Delta G=-541.4kJ/mol

Regards!

4 0
2 years ago
I2(g) + Cl2(g)2ICl(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings when 1.62 mol
galina1969 [7]

Answer:

The change in entropy of the surrounding is -146.11 J/K.

Explanation:

Enthalpy of formation of iodine gas = \Delta H_f_{(I_2)}=62.438 kJ/mol

Enthalpy of formation of chlorine gas = \Delta H_f_{(Cl_2)}=0 kJ/mol

Enthalpy of formation of ICl gas = \Delta H_f_{(ICl)}=17.78 kJ/mol

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H_{rxn}=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

For the given chemical reaction:

I_2(g)+Cl_2(g)\rightarrow 2ICl(g),\Delta H_{rxn}=?

The equation for the enthalpy change of the above reaction is:

\Delta H_{rxn}=[(2\times \Delta H_f_{(ICl)})]-[(1\times \Delta H_f_{(I_2)})+(1\times \Delta H_f_{(Cl_2)})]

=[2\times 17.78 kJ/mol]-[1\times 0 kJ/mol+1\times 62.436 kJ/mol]=-26.878 kJ/mol

Enthaply change when 1.62 moles of iodine gas recast:

\Delta H= \Delta H_{rxn}\times 1.62 mol=(-26.878 kJ/mol)\times 1.62 mol=-43.542 kJ

Entropy of the surrounding = \Delta S^o_{surr}=\frac{\Delta H}{T}

=\frac{-43.542 kJ}{298 K}=\frac{-43,542 J}{298 K}=-146.11 J/K

1 kJ = 1000 J

The change in entropy of the surrounding is -146.11 J/K.

4 0
3 years ago
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