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ludmilkaskok [199]

3 years ago

6

Which particle often described as having very little or no mass since the mass number does not change when you add more and more

of them to an atom?

1 answer:

astra-53 [7]3 years ago

7 0

Answer:

I wrote this answer to get points lol

Explanation:

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Which of the following statements is TRUE? Question 5 options: The emission spectrum of a particular element is always the same

amm1812

Answer:

All of the above are true

Explanation:

a) The emission spectrum of a particular element is always the same and can be used to identify the element: It's true since the emission spectrum for each element is unique. It has the same bright lines at the same wavelength. This feature is used to identify elements. For example, the study of the emission spectra of light arriving from stars allow us to identify the elements presents in the star because the light contains the emission spectra of those elements.

b)The uncertainty principle states that we can never know both the exact location and speed of an electron: It is true since the velocity of an electron is related to its wave nature, while its position is related to its particle nature and we cannot simultaneously measure electron's position and velocity with precision.

c) An orbital is the volume in which we are most likely to find an electron: An orbital is a probability distribution map that is used to decribe the likely position of an electron in an atom.

5 0

4 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

kumpel [21]

Answer:

Explanationrr

8 0

4 years ago

A piece of fossilized wood has a carbon-14 radioactivity that is 1/4 that of new wood. the half-life of carbon-14 is 5730 years.

Marizza181 [45]

Answer:

1.146 x 10⁴ year.

Explanation:

The decay of carbon-14 is a first order reaction.

The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.

The integration law of a first order reaction is:

<em>kt = ln [A₀]/[A]</em>

<em></em>

k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.

t is the time = ??? years.

[A₀] is the initial percentage of carbon-14 = 100.0 %.

[A] is the remaining percentage of carbon-14 = 1/4[A₀] = 25.0 %.

∵ kt = ln [Ao]/[A]

∴ (1.21 x 10⁻⁴ year⁻¹)(t) = ln (100.0%)/[25.0 %]

(1.21 x 10⁻⁴ year⁻¹)(t) = 1.386.

∴ <em>t </em>= 1.386/ (1.21 x 10⁻⁴ year⁻¹) = <em>1.146 x 10⁴ year.</em>

3 0

4 years ago

a vehicle of known mass accelerates along a straight path. according to newtons second law of motion, what caused this accelerat

postnew [5]

According to Newton's Second law of motion, t<span>he force acting on an object is equal to the mass of that object times its acceleration. The acceleration of the vehicle is caused by the force that is applied and the mass it posses. Hope this answers the question.</span>

8 0

4 years ago

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