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solniwko [45]
3 years ago
12

A sample of chlorine gas occupies a volume of 707 ml at 2.90 atm. What is the new pressure in torr if the volume is compressed t

o 0.533 L?
a: 3,850 torr
b: 3.85 torr
c: 291 torr
d: 2,920 torr
Chemistry
1 answer:
dolphi86 [110]3 years ago
3 0

Answer:

The new pressure is 3850 torr.  

Explanation:

The relation between volume and pressure is inverse as per Boyle's law. Its mathematical form is given by :

P_1V_1=P_2V_2

Here,

P_1=2.9\ atm

V_1=707\ mL

V_2=0.533\ L

Let P_2 is the new pressure. So using Boyle's law we get :

P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{2.9\times 707}{0.533}\\\\P_2=3846.71\ \text{torr}

or

P_2=3850\ \text{torr}

So, the new pressure is 3850 torr.  

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Answer:

The simplified expression for the fraction  is  \text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3 }

Explanation:

From the given information:

O3* → O3                   (1)    fluorescence

O + O2                      (2)    decomposition

O3* + M → O3 + M    (3)     deactivation

The rate of fluorescence = rate of constant (k₁) × Concentration of reactant (cO)

The rate of decomposition is = k₂ × cO

The rate of deactivation = k₃ × cO × cM

where cM is the concentration of the inert molecule

The fraction (X) of ozone molecules undergoing deactivation in terms of the rate constants can be expressed by using the formula:

\text {X} =    \dfrac{ \text {rate of deactivation} }{ \text {(rate of fluorescence) +(rate of decomposition) + (rate of deactivation) }  } }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{  {(k_1 \times cO) +(k_2 \times cO) + (k_3 \times cO \times cM) }  }

\text {X} =    \dfrac{  {k_3 \times cO \times cM} }{cO (k_1 +k_2 + k_3  \times cM) }

\text {X} =    \dfrac{  {k_3  \times cM} }{k_1 +k_2 + k_3  }    since  cM is the concentration of the inert molecule

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Answer:

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Explanation:

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nydimaria [60]

Answer:The correct answer is option (C).

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2. How many moles are in 8.30 x 1023 molecules of H2O?
daser333 [38]

Answer:

<h2>1.38 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

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From the question we have

n =  \frac{8.30 \times  {10}^{23} }{6.02 \times  {10}^{23} }  =  \frac{8.30}{6.02}  \\  = 1.3787 37...

We have the final answer as

<h3>1.38 moles</h3>

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8 0
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