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3 years ago

14

Benadryl, C17H21NO, is an over-the-counter drug used to treat allergy symptoms. If each dose of Benadryl contains 5.89 x 10^19mo

lecules of Benadryl, how many moles is this equal to?

1 answer:

Oksi-84 [34.3K]3 years ago

6 0

Answer:

9.78x10^(-5) mol

Explanation:

To convert molecules to moles, divide the # of molecules by 6.02x10^23.

5.89x10^19 / 6.02x10^23 = <u>9.78x10^(-5) mol</u>

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Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas.Express your a

ioda

Answer:

Explanation:

From the statement of the problem,

B₂S₃ $_{s}$ + H₂O $_{l}$ → H₃BO₃ $_{aq}$ + H₂S $_{g}$

B₂S₃ + H₂O → H₃BO₃ + H₂S

We that the above expression does not conform with the law of conservation of mass:

To obey the law, we need to derive a balanced reaction equation:

Let us use the mathematical method to obtain a balanced equation.

let the balanced equation be:

aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S

where a, b, c and d will make the equation balanced.

Conservating B: 2a = c

S: 3a = d

H: 2b = 3c + 2d

O: b = 3c

if a = 1,

c = 2,

b = 6,

2d = 2(6) - 3(2) = 6, d = 3

Now we can input this into our equation:

B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S

B₂S₃ $_{s}$ + 6H₂O $_{l}$ → 2H₃BO₃ $_{aq}$ + 3H₂S $_{g}$

4 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

What volume will be occupied by 5.00 mol of a gas at 25°C and 0.500<br> atm pressure?"

fomenos

Answer:

The volume will be occupied is 244, 36L.

Explanation:

We convert the unit of temperature to celsius into Kelvin, then use the ideal gas formula, solve for V (volume) and use the gas constant R =0.082 l atm / K mol:

0°C=273K 25°C= 273 + 25=298K

PV=nRT ---> V=nRT/P

V= 5,00 mol x 0,082 l atm/ K mol x 298 K/0,500 atm

<em>V=244,36L</em>

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3 years ago

How does the flow of energy during freezing compare with the flow of energy during melting

lianna [129]

Answer:

At the melting point, some molecules overcome the forces of attraction. Energy gained after the the solid melts increases the average kinetic energy or the temperature. When liquid is freezing the energy flows out the the liquid. As the kinetic energy decreases, they move more slowly.

Explanation:

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3 years ago

Multiplication & Division

Tatiana [17]

Answer:

0.008 ÷ 51.3 = 0.00015594541910331380.00015594541910331382Round 0.0001559454191033138 → 0.0002 (Sig Figs: 1)

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3 years ago

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