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cluponka [151]
3 years ago
9

The 64.8-g sample of the compound X2O5 contains 48.0 g of oxygen atoms. What is the molar mass of element X?

Chemistry
1 answer:
Vaselesa [24]3 years ago
8 0
The answer is 14 g/mol.

The number of moles (n) is the ratio of mass (m) and molar mass (M):
n = m : M

Step 1. Calculate number of moles of oxygen in the compound.
n = ?
m = 48 g
M = 16 g/mol

n = m/M = 48 g : 16 g/mol = 3 mol


Step 2. Calculate the number of moles of the element X:
5 atoms of oxygen is 3 mol.
2 atoms of element X is x
5 : 3  = 2 :x
x = 2 * 3 / 5 = 6/5 = 1.2 mol

Step 3. Calculate <span>the molar mass of element X
</span>n = 1.2 mol
m = 64.8 g - 48 g = 16.8 g
M = ?

n = m : M
M = m : n
M = 16.8 g : 1.2 mol = 14 g/mol
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Answer : The value of equilibrium constant for this reaction at 328.0 K is 1.70\times 10^{15}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = 151.2 kJ = 151200 J

\Delta S^o = standard entropy = 169.4 J/K

T = temperature of reaction = 328.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(151200J)-(328.0K\times 169.4J/K)

\Delta G^o=95636.8J=95.6kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = 95636.8 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 328.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

95636.8J=-(8.314J/K.mol)\times (328.0K)\times \ln k

k=1.70\times 10^{15}

Therefore, the value of equilibrium constant for this reaction at 328.0 K is 1.70\times 10^{15}

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3 years ago
Which discovery did J. J. Thomson make that improved upon Dalton's atomic theory?
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Answer: Atoms contain tiny, negatively charged electrons

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Why do we say the particles in a rock lying on the ground have kinetic energy and potential energy?
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Explanation:

hop it helps

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3 years ago
slader Consider the following reactions: A. uranium-238 emits an alpha particle; B. plutonium-239 emits an alpha parti- cle; C.
pishuonlain [190]

<u>Answer:</u>

<u>For A:</u> The equation is _{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> The equation is _{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> The equation is _{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

<u>Explanation:</u>

Alpha decay process is the process in which nucleus of an atom disintegrates into two particles. The first one which is the alpha particle consists of two protons and two neutrons. This is also known as helium nucleus. The second particle is the daughter nuclei which is the original nucleus minus the alpha particle released.

_Z^A\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\alpha

Beta decay process is defined as the process the neutrons get converted into an electron and a proton. The released electron is known as the beta particle. In this process, the atomic number of the daughter nuclei gets increased by a factor of 1 but the mass number remains the same.

_Z^A\textrm{X}\rightarrow _{Z+1}^A\textrm{Y}+_{-1}^0\beta

<u>For A:</u> Uranium-238 emits an alpha particle

The nuclear equation for this process follows:

_{92}^{238}\textrm{U}\rightarrow _{90}^{234}\textrm{Th}+_2^4\alpha

<u>For B:</u> Plutonium-239 emits an alpha particle

The nuclear equation for this process follows:

_{94}^{239}\textrm{Pu}\rightarrow _{92}^{235}\textrm{U}+_2^4\alpha

<u>For C:</u> Thorium-239 emits a beta particle

The nuclear equation for this process follows:

_{90}^{239}\textrm{Th}\rightarrow _{91}^{235}\textrm{Pa}+_{-1}^0\beta

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What does resonance result in?
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louder sounds

Explanation:

Resonance results in louder sounds.

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Resonance results in sound amplification and causes louder sounds. Increasing vibration results in the production of a louder sound.

Learn more:

Resonance in chemistry brainly.com/question/9422880

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