Please help!!!!! The question is in the answer..

Complete the statement qxy- bxy+cxy= xy( )

andrey2020 [161]

Answer:

xy (-b+c+q) is the answer to this

3 0

3 years ago

When a bullet is fired from a gun what is true about the momentum

Eduardwww [97]

Before the bullet is fired the momentum is Zero because nothing is moving but once the bullet is shot the momentum increases because of the movement of the bullet moving forward.

8 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]

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

Explanation:

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

Hi. I need help with this

Vlad1618 [11]

Factors that determine ionization energy:

Electronic Repulsion - If the electronic density decreases, the ionization energy with increase and vice versa. If an electron gets released, it decreases the electronic repulsion. This makes releasing another electron harder than the first on unless the electron that is being released comes from another energetic level.
# Of Energy Levels - The more energy cores that get filled up, the more ionization levels decrease. When we see the energy levels go from top to bottom, the ionization also go from most to least. This is why ionization occurs on the highest level.
Nuclear Charge - The higher the atomic number, the higher charge in the nuclei. This also makes the ionization energy higher as it increases from left to right of in other words, if the nuclear charge gets higer, the energy gets higher as well.

Factors that determine atomic volume:

How many protons are in the nucleus (nuclear charge)
How many energy levels carry electrons (electrons in outer energy level)

Best of Luck!

4 0

3 years ago

PLEASE HELP. Give the orbital configuration of the phosphorus (p) atom.

Kipish [7]

Answer: 1s^22s^22p^63s^23p^3

Explanation:

Assuming that orbital configuration is the same as electron configuration this is the answer.

4 0

3 years ago

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