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3 years ago

9

in this cut-away image of an atom, where is the red arrow most likely pointed at the greatest probability of locating an electro

n

2 answers:

iren [92.7K]3 years ago

5 0

It’s C :) hopefully its right ^_^

emmasim [6.3K]3 years ago

3 0

Answer:

c

Explanation:

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In Denver, Colorado the elevation is about 5,280 feet above sea level. Explain what potential effects this may have on the solub

Colt1911 [192]

Answer:

The solubility of the gaseous solute decreases

Explanation:

As we know, pressure decreases with altitude. This means that, at higher altitudes, the pressure is much lower than it is at sea level.

The solubility of a gas increases with increase in pressure and decreases with decrease in pressure.

Hence, in Denver, Colorado where the elevation is about 5,280 feet above sea level, a gaseous solute is less soluble than it is at sea level due to the lower pressure at such high altitude.

4 0

3 years ago

The chemical equation for a reaction between K2Cr2O7 and HCl is shown. K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O

guapka [62]

<u>Answer: </u>The correct answer is Option b.

<u>Explanation:</u>

Oxidizing agent is defined as the chemical reagent which helps the other chemical compound to get oxidized and itself gets reduced. The oxidation state for these species gets reduced because they are undergoing reduction reaction.

For the given chemical equation:

$K_2Cr_2O_7+14HCl\rightarrow 2CrCl_3+2KCl+3Cl_2+7H_2O$

Oxidation state of Chromium is getting reduced from +6 to +3 and oxidation state of chlorine getting increased from -1 to 0.

Hence, $K_2Cr_2O_7$ acts like and oxidizing agent because it is itself getting reduced to $CrCl_3$

Therefore, the correct answer is Option b.

4 0

3 years ago

A solution contains 15.27 grams of NaCl in 0.670 kg water at 25 °C. What is the vapor pressure of the solution?

Anvisha [2.4K]

Answer:

The vapor pressure of the solution is 23.636 torr

Explanation:

$P_{solution} = X_{solvent}*P_{solvent}$

Where;

$P_{solution$ is the vapor pressure of the solution

$X_{solvent$ is the mole fraction of the solvent

$P_{solvent$ is the vapor pressure of the pure solvent

Thus,

15.27 g of NaCl = [(15.27)/(58.5)]moles = 0.261 moles of NaCl

0.67 kg of water = [(0.67*1000)/(18)]moles = 37.222 moles of H₂O

Mole fraction of solvent (water) = (number of moles of water)/(total number of moles present in solution)

Mole fraction of solvent (water) = (37.222)/(37.222+0.261)

Mole fraction of solvent (water) = 0.993

<u>Note:</u> the vapor pressure of water at 25°C is 0.0313 atm

Therefore, the vapor pressure of the solution = 0.993 * 0.0313 atm

the vapor pressure of the solution = 0.0311 atm = 23.636 torr

5 0

3 years ago

Read 2 more answers

7. A small stone added to boiling liquids to make them boil more evenly without sudden violent releases of

Rzqust [24]

Answer:

A boiling chip, boiling stone, porous bit or anti-bumping granule is a tiny, unevenly shaped piece of substance added to liquids to make them boil more calmly.

These help in making the liquid boil more easily

8 0

3 years ago

Read 2 more answers

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