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Eddi Din [679]
2 years ago
6

How long would it take for 1.50 mol of water at 100.0 ∘c to be converted completely into steam if heat were added at a constant

rate of 19.0 j/s ?
Chemistry
2 answers:
blagie [28]2 years ago
6 0

\boxed{{\text{3213}}{\text{.15 s}}} will be required to convert 1.50 mol of water completely into steam.

Further explanation:

Enthalpy of vaporization

It is the amount of energy that is required to convert a substance from its liquid state to a vapor or gaseous state. It is also known as the latent heat of vaporization or heat of evaporation. It is represented by \Delta {H_{{\text{vap}}}}.

The expression for the heat of vaporization is as follows:

{\text{q}} = {\text{n}}\Delta {H_{{\text{vap}}}}                                                                …… (1)

Here,

q is the energy of the substance.

n is the number of moles of the substance.

\Delta {H_{{\text{vap}}}} is the heat of vaporization.

Substitute 1.50 mol for n and 40.7 kJ/mol for \Delta {H_{{\text{vap}}}} in equation (1).

 \begin{aligned}{\text{q}}&= \left( {1.50{\text{ mol}}} \right)\left( {\frac{{40.7{\text{ kJ}}}}{{1{\text{ mol}}}}} \right)\\&= 61.05{\text{ kJ}}\\\end{aligned}

The amount of energy is to be converted into J. The conversion factor for this is,

 1{\text{ kJ}} = {\text{1}}{{\text{0}}^3}{\text{ J}}

Therefore the amount of energy can be calculated as follows:

\begin{aligned}{\text{q}} &= \left( {61.05{\text{ kJ}}} \right)\left( {\frac{{{{10}^3}{\text{ J}}}}{{1{\text{ kJ}}}}} \right)\\&= {\text{61050 J}}\\\end{aligned}

The time required for conversion of water into steam is calculated as follows:

\begin{aligned}{\text{Time required}}&= \left( {61050{\text{ J}}} \right)\left( {\frac{{1{\text{ s}}}}{{19{\text{ J}}}}} \right)\\&= 3213.15{\text{ s}}\\\end{aligned}  

Learn more:

  1. Calculate the enthalpy change using Hess’s Law: brainly.com/question/11293201
  2. Find the enthalpy of decomposition of 1 mole of MgO: brainly.com/question/2416245

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Thermodynamics

Keywords: enthalpy of vaporization, q, n, 1.50 mol, water, steam, 3213.15 s, 61050 J, 61.05 kJ, 40.7 kJ/mol, liquid state, vapour state, substance, time, amount of energy.

Aleksandr-060686 [28]2 years ago
5 0
To determine the time it takes to completely vaporize the given amount of water, we first determine the total heat that is being absorbed from the process. To do this, we need information on the latent heat of vaporization of water. This heat is being absorbed by the process of phase change without any change in the temperature of the system. For water, it is equal to 40.8 kJ / mol.

Total heat = 40.8 kJ / mol ( 1.50 mol ) = 61.2 kJ of heat is to be absorbed

Given the constant rate of 19.0 J/s supply of energy to the system, we determine the time as follows:

Time = 61.2 kJ ( 1000 J / 1 kJ ) / 19.0 J/s = 3221.05 s
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Answer:

Inventory management documentation

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Inventory management documentation will surely aid Lin to know what employee has received his/her laptop and who has not from his inventory records. The scope of Inventory management documentation transverses the entire spectrum of a supply chain.

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What is the maximum volume of a 0.788 M CaCl2 solution that can be prepared using 85.3 g CaCl2?
Anna11 [10]
Molar mass  CaCl₂ =  110.98 g/mol

Number of moles:

1 mole CaCl₂ ---------> 110.98 g
n mole CaCl2 ---------> 85.3 g

n = 85.3 / 110.98

n = 0.7686 moles of CaCl₂

Volume = ?

M = n / V

0.788 =  0.7686 / V

V = 0.7686 / 0.788

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2 years ago
How many moles of chromium(iii) nitrate are produced when chromium reacts with 0.85 moles of lead(iv) nitrate to produce chromiu
Alexxandr [17]
The  moles  of  chromium (iii)  nitrate  produced  is  calculated  as   follows

write  the  equation  for  reaction

 3  Pb(NO3)2  +  2 Cr  =  2 Cr(NO3)3  +  3  Pb

by  use  of  mole  ratio  between  Pb(NO3)2  to  Cr(NO3)3  which  is  3  :  2  the  moles  of  Cr(NO3)3  is therefore  
=  0.85  x2  /3  =  0.57   moles
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A 0.2-mm-thick wafer of silicon is treated so that a uniform concentration gradient of antimony is produced. One surface contain
krek1111 [17]

Answer:

- 0.0249% Sb/cm

-1.2465 * 10^9 \frac{atoms}{cm^3.cm}

Explanation:

Given that:

One surface contains 1 Sb atom per  10⁸  Si atoms and the other surface contains 500 Sb atoms per  10⁸ Si atoms.

The concentration gradient in atomic percent (%) Sb  per cm can be calculated as follows:

The difference in concentration = \delta_c

The distance \delta_x = 0.2-mm = 0.02 cm

Now, the concentration of silicon at one surface containing  1 Sb atom per 10⁸ silicon atoms and at the outer surface that has 500 Sb atom per   10⁸ silicon atoms can be calculated as follows:

\frac{\delta_c}{\delta_c} = \frac{(1/10^8 -500/10^8)}{0.02cm} *100%

= - 0.0249% Sb/cm

b) The concentration (c_1) of Sb in atom/cm³ for the surface of 1 Sb atoms can be calculated by using the formula:

c_1 = \frac{(8 si atoms/unit cells)(1/10^3)}{(lattice parameter)^3/unit cell}

Lattice parameter = 5.4307 Å;  To cm ; we have

= 5.4307A^0* \frac{10^{-8}cm}{ A^0}

c_1 = \frac{(8 si atoms/unit cells)(1/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

= 0.00499*10^{17}atoms/cm^3

The concentration (c_2) of Sb in atom/cm³ for the surface of 500 Sb can be calculated as follows:

c_1 = \frac{(8 si atoms/unit cells)(500/10^8)}{(5.4307*10^{-8}cm)^3/unit cell}

   =  \frac{4*10^{-3}}{1.601*10^{-22}}

   = 2.4938*10^{17}atoms/cm^3

Finally, to calculate the concentration gradient

(\frac{\delta _c}{\delta_ x}) = \frac{c_1-c_2}{\delta_x}

(\frac{\delta _c}{\delta_ x}) = \frac{0.00499*10^{17}-2.493*10^{17}}{0.02}

= -1.2465 * 10^9 \frac{atoms}{cm^3.cm}

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