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julsineya [31]
3 years ago
13

The water bottle contains 575 grams of water at 80°C. The water eventually cools to

Chemistry
1 answer:
Jet001 [13]3 years ago
8 0

Answer:

–23000 Calories.

NOTE : The negative sign indicates that heat has been loss to the student back.

Explanation:

The following data were obtained from the question:

Mass (M) of water = 575 g

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Heat (Q) transferred =.?

Next, we shall determine the change in temperature (ΔT).

This is illustrated below:

Initial temperature (T1) = 80 °C

Final temperature (T2) = 40 °C

Change in temperature (ΔT) =?

Change in temperature (ΔT) = T2 – T1

Change in temperature (ΔT) = 40 – 80

Change in temperature (ΔT) = –40°C

Finally, we shall determine the heat transferred. This can be obtained as follow:

Mass (M) of water = 575 g

Change in temperature (ΔT) = –40°C

Specific heat capacity (C) of water = 1 Cal/g°C

Heat (Q) transferred =.?

Q = MCΔT

Q = 575 × 1 × –40

Q = –23000 Calories

NOTE : The negative sign indicates that heat has been loss to the student back.

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Which of the following statements best describes what happens to water during vaporization?
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Answer: im pretty sure its D

Explanation:

8 0
3 years ago
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A chemist fills a reaction vessel with 9.20 atm nitrogen monoxide (NO) gas, 9.15 atm chlorine (CI) gas, and 7.70 atm nitrosyl ch
ivanzaharov [21]

Answer:

The reactions free energy \Delta G = -49.36 kJ

Explanation:

From the question we are told that

      The pressure of (NO) is P_{NO} = 9.20 \ atm

      The  pressure of  (Cl) gas is  P_{Cl} = 9.15 \ atm

       The  pressure of nitrosly chloride (NOCl) is P_{(NOCl)} = 7.70 \ atm

The reaction is

              2NO_{(g)} + Cl_2 (g)    ⇆   2 NOCl_{(g)}

 From the reaction we can  mathematically evaluate the \Delta G^o (Standard state  free energy ) as

                    \Delta G^o = 2 \Delta G^o _{NOCl} -   \Delta G^o _{Cl_2}  - 2 \Delta G^o _{NO}

The Standard state  free energy for NO is  constant with a value  

                 \Delta G^o _{NO} = 86.55 kJ/mol

 The Standard state  free energy for Cl_2 is  constant with a value                  

             \Delta G^o _{Cl_2} = 0kJ/mol

 The Standard state  free energy for NOCl is  constant with a value

         \Delta G^o _{NOCl} =66.1kJ/mol

Now substituting this into the equation

        \Delta G^o = 2 * 66.1 - 0 - 2 * 87.6

                = -43 kJ/mol

The pressure constant is evaluated as

         Q =  \frac{Pressure \ of  \ product }{ Pressure  \ of \ reactant }

Substituting  values  

        Q = \frac{(7.7)^2 }{(9.2)^2 (9.15) } = \frac{59.29}{774.456}

           = 0.0765

The free energy for this reaction is evaluated as

           \Delta  G  =  \Delta  G^o  + RT ln Q

Where R is gas constant with a value  of  R = 8.314 J / K \cdot mol

          T is temperature in K  with a given value of  T = 25+273 = 298 K

   Substituting value

                \Delta  G  = -43 *10^{3} + 8.314 *298 * ln [0.0765]

                       = -43-6.36

                      \Delta G = -49.36 kJ

4 0
3 years ago
When a substance is in the liquid state, how are the particles of that substance<br> behaving
iVinArrow [24]

Answer:

In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles. With an increase in temperature, the particles move faster as they gain kinetic energy, resulting in increased collision rates and an increased rate of diffusion.

Explanation:

In liquids, particles are quite close together and move with random motion throughout the container. Particles move rapidly in all directions but collide with each other more frequently than in gases due to shorter distances between particles. With an increase in temperature, the particles move faster as they gain kinetic energy, resulting in increased collision rates and an increased rate of diffusion.

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3 years ago
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What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
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Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

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The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

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How many moles of HCI would be required to produce a total of 2 moles of H2
k0ka [10]

Answer:

Should be 4 moles.

4 0
3 years ago
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