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allochka39001 [22]
3 years ago
14

How is gas different from a liquid

Chemistry
2 answers:
vlada-n [284]3 years ago
6 0
The gas is different from a liquid because the gas can escape a liquid doesn't
satela [25.4K]3 years ago
4 0
The atoms and molecules in gases<span> are much more spread out than in solids or</span>liquids<span>. They vibrate and move freely at high speeds. A </span>gas<span> will fill any container, but if the container is not sealed, the </span>gas<span> will escape. </span>Gas<span> can be compressed much more easily than a </span>liquid<span> or solid</span>
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PLEASE HELP!!!!!!!!!! DOING A TEST
Nastasia [14]
D is the correct answer
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3 years ago
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The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the
IceJOKER [234]

Answer : The concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

Explanation :

The expression for first order reaction is:

[C_t]=[C_o]e^{-kt}

where,

[C_t] = concentration at time 't'  (final) = ?

[C_o] = concentration at time '0' (initial) = 0.100 M

k = rate constant = 5.40\times 10^{-3}s^{-1}

t = time = 17.0 min = 1020 s (1 min = 60 s)

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

[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}

[C_t]=4.05\times 10^{-4}M

Thus, the concentration after 17.0 minutes will be, 4.05\times 10^{-4}M

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3 years ago
Never mind jhvjycdtrsesetdfyguhbjnk
OlgaM077 [116]

Complete Question

methanol can be synthesized in the gas phase by the reaction of gas phase carbon monoxide with gas phase hydrogen, a 10.0 L reaction flask contains carbon monoxide gas at 0.461 bar and 22.0 degrees Celsius. 200 mL of hydrogen gas at 7.10 bar and 271 K is introduced. Assuming the reaction goes to completion (100% yield)

what are the partial pressures of each gas at the end of the reaction, once the temperature has returned to 22.0 degrees C express final answer in units of bar

Answer:

The partial  pressure of  methanol is  P_{CH_3OH_{(g)}} =0.077 \  bar

The partial  pressure of carbon monoxide is  P_{CO} = 0.382 \ bar

The partial  pressure at  hydrogen is  P_H =  O \  bar

Explanation:

From the question we are told that

  The volume of the  flask is  V_f = 10.0 \  L

   The initial pressure of carbon monoxide gas is  P_{CO} = 0.461 \ bar

   The initial  temperature of carbon monoxide gas is T_{CO} = 22.0^oC

   The volume of the hydrogen gas is  V_h  =  200 mL = 200 *10^{-3} \  L

    The initial  pressure of the hydrogen is P_H  =  7.10 \  bar

    The initial temperature of the hydrogen  is  T_H = 271 \  K

The reaction of  carbon monoxide and  hydrogen is  represented as

         CO_{(g)} + 2H_2_{(g)} \rightarrow CH_3OH_{(g)}

Generally from the ideal gas equation the initial number of moles of carbon monoxide is  

        n_1  =  \frac{P_{CO} *  V_f }{RT_{CO}}

Here R is the gas constant with value  R  = 0.0821 \ L \cdot atm \cdot mol^{-1} \cdot K

=>     n_1  =  \frac{0.461  *  10 }{0.0821 * (22 + 273)}

=>     n_1  = 0.19

Generally from the ideal gas equation the initial number of moles of Hydrogen  is  

       n_2  =  \frac{P_{H} *  V_H }{RT_{H}}

      n_2  =  \frac{ 7.10 *  0.2 }{0.0821 * 271 }

=> n_2  =  0.064

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CO

=>      0.064 moles of  hydrogen gas will react with  x  mole of  CO

So

          x = \frac{0.064}{2}

=>       x = 0.032 \ moles \ of  \  CO

Generally from the chemical equation of the reaction we see that

        2 moles of hydrogen gas reacts with 1 mole of CH_3OH_{(g)}

=>      0.064 moles of  hydrogen gas will react with  z  mole of  CH_3OH_{(g)}

So

          z = \frac{0.064}{2}

=>       z = 0.032 \ moles \ of  \ CH_3OH_{(g)}

From this calculation we see that the limiting reactant is hydrogen

Hence the remaining CO after the reaction is  

          n_k = n_1 - x

=>       n_k = 0.19  - 0.032

=>       n_k = 0.156

So at the end of the reaction , the partial pressure for  CO is mathematically represented as

      P_{CO} = \frac{n_k  *  R *  T_{CO}}{V}

=>    P_{CO} = \frac{0.158   *  0.0821 *  295}{10}

=>    P_{CO} = 0.382 \ bar

Generally the partial pressure of  hydrogen is  0 bar because hydrogen was completely consumed given that it was the limiting reactant

Generally the partial  pressure of the methanol is mathematically represented as

         P_{CH_3OH_{(g)}} = \frac{z  *  R *  T_{CO}}{V_f}

Here  T_{CO} is used because it is given the question that the   temperature  returned to 22.0 degrees C

So

      P_{CH_3OH_{(g)}} = \frac{0.03 * 0.0821 *  295}{10}

     P_{CH_3OH_{(g)}} =0.077 \  bar

6 0
3 years ago
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In the laboratory you dissolve 15.9 g of barium chloride in a volumetric flask and add water to a total volume of 375 ml. what i
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<span>There are a number of ways to express concentration of a solution. This includes molarity. Molarity is expressed as the number of moles of solute per volume of the solution. So, we calculate as follows:

Molarity =  15.9 g BaCl2 ( 1 mol / 208.23 g ) / .375 L = 0.204 mol / L</span>
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3 years ago
I am having trouble converting :(
Elza [17]

Answer:

For the first question, to determine the total number of molecules of nitrogen dioxide, first make use of the molar mass of the nonpolar compound and then use that to find the total number of moles and then subsequently after make use of the ratio for the Avogadro's number to determine the total number of molecules of this compound.

For the final question, do the inverse, where we make use of the molecules of the compound and then use Avogadro's number to determine the moles of the compound and then use the same molar mass of the compound to determine the grams of the Nitrogen Dioxide.

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3 years ago
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