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julia-pushkina [17]
3 years ago
12

Give two characteristics of liquid state​

Chemistry
1 answer:
inna [77]3 years ago
8 0

Answer:  Liquids have the following characteristics: no definite shape (takes the shape of its container) has definite volume. particles are free to move over each other, but are still attracted to each other. liquids have lesser densities than solids.  Intermolecular forces of attraction is weaker than solids.  They have considerable space between the particles. Liquids flow from higher to lower level.  Liquids have their boiling points above room temperature, under normal conditions.

I gave many characteristics

Hope this helps :)

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Which substance is the oxidizing agent in the following reaction? Fe2S3 + 12HNO3 → 2Fe(NO3)3 + 3S + 6NO2 + 6H2O Which substance
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Answer:

HNO3 is the oxidizing agent

Explanation:

Step 1:

The oxidizing agent is responsible to oxidize another.  but  itself undergoes  a reduction  .

Fe2S3:

Fe has an oxidation number of +3

S has an oxidation number of -2

HNO3:

O has an oxidation number of -2 ( we have 3 times O so this makes -6)

H has an oxidation number of +1

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O has an oxidation number of -2 ( we have 3 times O so this makes -6)

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S alone has an oxidation number of 0

NO2:

O has an oxidation number of -2 (we have 2 times O, this makes -4)

N has an oxidation number of +4

Fe doesn't change from oxidation number. It stays +3

N goes from +5 to +4 → this is a reduction

S goes from -2 to 0 → this is an oxidation

The reducing agent is the compound that contributes the oxidized species (S).

The oxidizing agent contributes the reducing species (N)

The answer is HNO3

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Which of the following is a reasonable ground-state electron configuration?
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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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