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3 years ago

15 points will mark brainliest

Chemistry

2 answers:

dybincka [34]3 years ago

6 0

Answer:

Option b as 100 years before also rain was not there is sahara dessert and now also it is not there..

Vladimir [108]3 years ago

4 0

You got it right, the answer is d

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3 years ago

Why do most atoms form chemical bonds?

zloy xaker [14]

Answer:

Why do most atoms form chemical bonds? They want a full outer shell of electrons, so the lose, gain, or share electrons with other elements, forming compounds, until they have 8 valence electrons and become stable. Double and triple covalent bonds that have greater bond energy and are shorter than single bonds.

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3 years ago

How are the products of the endoplasmic reticulum shipped to the golgi?

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The ribosomes are the ones delivering the products of the endoplasmic reticulum

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3 years ago

Uranium has atomic number 92. Its most common isotope is 238U, but the form used in nuclear bombs and nuclear power plants is 23

Yakvenalex [24]

Explanation:

Atomic Number = Number of protons

Mass Number = Number of protons + Number of neutrons

Isotopes are simply atoms of an element with the same number of protons and different number of neutrons.

First Isotope -- 238U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons = 238 - 92 = 146

Second Isotope -- 235U

Number of neutrons = Mass Number - Atomic Number

Number of neutrons = 235 - 92 = 143

3 0

3 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