How do ionization energies provide evidence for the quantization of the energies of electrons?

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

vichka [17]

Answer:

$\Delta G =-103.95kJ$

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

$\Delta G =\Delta H-T\Delta S$

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

$\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ$

Best regards!

6 0

3 years ago

1. The following reaction does not proceed to form a product: H2O + Au---> no reaction. Why is that?

Xelga [282]

1. B
The positive charge in water is provided by hydrogen, and gold provides the same charge. However, gold is not more reactive than hydrogen so it can not replace it in the compound.

2. In order to balance the equation, you must sure there are equal moles of each element on the left and right side of the equation:
2C₂H₆ + 7O₂ → 4CO₂ + ₆H₂O

3. The number of moles of sodium atoms on the left of the equation must be equal to the number of moles of sodium atoms on the right, as per the law of conservation of mass. The answer is B.

4. C.
A synthesis reaction usually results from single displacement because some element or compound is produced in its pure form

5. B.
The gas being produced is being synthesized.

8 0

3 years ago

Read 2 more answers

Please answer number 2

alexandr402 [8]

Answer:

A

Explanation:

Because Rainwater can get acidic because of the carbonic acid that it contains

8 0

3 years ago

I need help with number 8 please help ASAP.

Rzqust [24]

Answer:

33.33% = 33%

Explanation:

MgCO3(s) + 2HCl (aq) --> MgCl2(aq) + H20(l) + CO2(g)

1 mole of MCO3 will produce → 1 mole of CO2

We need to get the number of mole of CO2:

and when we have 0.22 g of CO2, so number of mole = mass / molar mass

Moles = 0.22 g / 44 g/mol = 0.005 mole

Moles of Mg = moles of CO2 = 0.005 mole

Mass of Mg = moles * molar mass

= 0.005 * 84 /mol = 0.42 g

Percent of MgCO3 by mass of Mg = 0.42 g / 1.26 * 100

=33.33 %

8 0

3 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0

3 years ago