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3 years ago

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The kinetic-molecular theory of gases says A. the particles of a gas move independently of each other. B. the particles in a gas

move rapidly. C. the particles in a gas are far apart. D. all of the above.

1 answer:

Korolek [52]3 years ago

5 0

Answer:

D. all of the above.

Explanation:

The kinetic-molecular theory of gases suggests that; the particles of a gas move independently of each other, the particles in a gas move rapidly, the particles in a gas are far apart.

Hope it helps.

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Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S_f^0$ = standard entropy of formation

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (189J/K.mol)+2mole\times (248J/K.mol)}]-[2mole\times (206J/K.mol)+3mole\times (205J/K.mol)]$

$\Delta S^o=-153J/K$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is 500 K.

$\Delta G^o=(-1035600J)-(500K\times -153J/K)$

$\Delta G^o=-959100J=-959.1kJ$

Therefore, the value of $\Delta G^o$ for the reaction is -959.1 kJ

3 0

3 years ago

- How much power does it take to lift a<br> 1,000 N load 10 m in 20 s?

Mariana [72]

Answer:

"500 Joule/sec" is the right answer.

Explanation:

The given values are:

Force,

F = 1000 N

Velocity,

s = 10 m

Time,

t = 20 s

Now,

The power will be:

= $\frac{Force\times Velocity}{Time}$

On putting the values, we get

= $\frac{1000\times 10}{20}$

= $\frac{10000}{20}$

= $500 \ Joule/sec$

5 0

3 years ago

What are some insulators and conductors of electricity? Name 2 of each

hammer [34]

Answer:

insulator:wood,rubber

conductor:aluminium,copper

4 0

2 years ago

If 26.2 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.785-mg sample of KCl (forming AgCl), what is the molarity o

sweet-ann [11.9K]

<u>Answer:</u>

<em>The molarity of the $AgNO_3$ solution is $4.02 \times 10^4 M$ </em>

<em></em>

<u>Explanation:</u>

The Balanced chemical equation is

$1AgNO_3 (aq) +1KCl (aq) > 1 AgCl (s)+1KNO_3 (aq)$

Mole ratio of $AgNO_3$ : KCl is 1 : 1

So moles $AgNO_3$ = moles KCl

$Moles KCl = \frac {mass}{molarmass}$

$= \frac {0.785 mg}{(39.1+35.5 g per mol)}$

$= \frac {0.000785 g}{74.6 g per mol}$

$= 0. 0000105 mol KCl$

$= 0.0000105 mol AgNO_3$

So Molarity

$= \frac {moles of solute}{(volume of solution in L)}$

$= \frac {0.0000105 mol}{26.2 mL}$

$=\frac {0.0000105 mol}{0.0262 L}$

= 0.000402M or mol/L is the Answer

(Or) $4.02 \times 10^4 M$ is the Answer

6 0

3 years ago

Someone please help! this is the last question<br>I only need help with B.<br><br>

ludmilkaskok [199]

1. mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. Na₂CO₃ as a limiting reactant

<h3>Further explanation</h3>

Given

Reaction

2 Al(NO₃)₃ + 3 Na₂CO₃ → Al₂(CO₃)₃ + 6 NaNO₃

Required

mol ratio

Limiting reactant

Solution

The reaction coefficient in the chemical equation shows the mole ratio of the components of the compound involved in the reaction (reactants and products)

1. From the equation mol ratio of Al(NO₃)₃ : Na₂CO₃ = 2 : 3

2. mol : coefficient of Al(NO₃)₃ : Na₂CO₃ = 2 mole/2 : 2 mole/3 = 1 : 0.67

Na₂CO₃ as a limiting reactant (smaller)

6 0

2 years ago