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2 years ago

How much heat is required to convert 422 g of liquid h2o at 23.5 °c into steam at 150 °c?

2 answers:

4 0

Heat is given by multiplying the specific heat capacity of a substance by mass and the change in temperature. The heat capacity of water is Approximately 4184 J/K/C.
Therefore, heat = mc0 mass in kg
= (422/1000) × 4184 × (100-23.5)
= 135072.072 J
Latent heat of vaporization is 2260 kJ/kg
Thus the heat will be 0.422 × 2260000 = 953720 J
Heat to raise steam from 100 to 150
2000 × 0.422 ×50 = 42200 J
Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules or 1330 kilo joules

erik [133]2 years ago

3 0

The heat that is required to convert 422 g of liquid H₂O at 23.5 °c into steam at 150 °c can be calculated as follows:

The heat capacity of water = 4184 J K⁻¹ C⁻¹

Therefore, heat required to warm the water from

23.5 °C to 100.0 °C.

= $m\times c\times \Delta T$

Here,

m=0.422 kg

c=4184 J K⁻¹ C⁻¹

$\Delta T$ =100-23.5

so, heat required to warm the water from 23.5 °C to 100.0 °C

= 0.422 × 4184 × (100-23.5)

= 135072.072 J

Latent heat of vaporization of water is 2260 kJ/kg

Thus the heat will be 0.422 × 2260000 = 953720 J

Heat required to raise steam from 100 to 150

2000 × 0.422 ×50 = 42200 J

Thus the heat required is (135072.072 + 953720 + 42200) = 1330992.07 Joules

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MrMuchimi

Answer:

El volumen del cuerpo es el mismo al comienzo de la experiencia.

Explicación:

El volumen del cuerpo es el mismo al principio porque el volumen no cambia si la temperatura permanece igual. Si cambiamos la temperatura i. mi. Al aumentar la temperatura, las moléculas comienzan a expandirse y se produce un aumento de volumen mientras que cuando disminuimos la temperatura, las moléculas de esa sustancia comienzan a contraerse y el volumen de esa sustancia disminuye. Entonces concluimos que el volumen depende de la temperatura.

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1 year ago

What volume of 0.210 M sulfuric acid is required to completely react with 2.14 g aluminium hydroxide?

Galina-37 [17]

3 H2SO4 + 2 Al(OH)3 → Al2(SO4)3 + 6 H2O

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3 0

1 year ago

The concentration of C29H60 in summer rainwater is 34 ppb. Find the molarity of this compound in nanomoles per liter (nM).

Naya [18.7K]

Answer: The concentration of C29H60 in nM per liter is 83,33 nM/liter

Explanation: Let's start from the ppb definition: ppb means parts per billion. In terms of concentracion measuring this means micrograms of solute per liter of solution.

The algebraic expression would be:

<em>ppb [=] micrograms of compound/liter of solution</em>

We can assume that the solvent is water. The solute is dissolved in water and both create the C29H60 solution.

For the exercise we have 34 ppb of C29H60, that means 34 micrograms of C29H60 in one liter of solution. So, since now, we have to convert the units from the initial data to the required answer.

The respective procedure is in a attached file.

5 0

1 year ago

1. A student performs a reaction in a beaker by reaction silver nitrate with one long copper wire. The student’s thermometer cha

stiv31 [10]

Explanation:

The observation of student was that thermometer reading changed from 27°C to 35°C which indicates that temperature of the beaker solution rose after reaction due to release of heat during reaction as a product.

Those chemical reactions which gives heat energy as a product into their surrounding are categorized as exothermic reactions. During the course of these reaction temperature of the surroundings also increased.

So, this means that reaction between silver nitrate and copper wire is an exothermic reaction.

Three ways that the student could speed up the reaction :

By adding catalyst to the reaction.
By decreasing the temperature.
By increasing the concentration of silver nitarte solution.

3 0

2 years ago

Read 2 more answers

6. Under standard-state conditions, what spontaneous reaction will occur in aqueous solution among the ions Ce4+, Ce3+, Fe3+, an

xz_007 [3.2K]

Answer:

ΔG° = -80.9 KJ

Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

Explanation:

1) Reduction potentials

First of all one should look up the reduction potentials for the species envolved:

$Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

$Fe^{3+} + e$ → $Fe^{2+}$ E°red=0.771V

2) Redox pair

Knowing their reduction pontentials one can determine a redox pair: one species must oxidate while the other is reducing. <u>Remember: the table gives us the reduction potential, so if we want to know the oxidation potential all that has to be done is reverce the equation and change the potencial signal (multiply to -1).</u>

1) $Ce^{4+}$ reduces while $Fe^{2+}$ oxidates

(oxidation) $Fe^{2+}$ → $Fe^{3+} + e$ E°oxi=-0.771V

(reduction) $Ce^{4+} + e$ → $Ce^{3+}$ E°red=1.61V

(overall equation) $Fe^{2+}+Ce^{4+}$ → $Ce^{3+}+Fe^{3+}$ E°=Ereduction + Eoxidation= 1.61 v+(-0.771 v) = 0.839v

The cell potential can also be calculated as the cathode potencial minus the anode potential:

E° = E cathode - E anode =1.61 v - 0.771 v=0.839 v

3) Gibbs free energy and Equilibrium constant

ΔG°=-nFE°, where 'n' is the number of electrons involved in the redox equation, in this case n is 1. 'F' is the Faraday constant, whtch is 96500 C. E° is the standard cell potencial.

ΔG°=-nFE°=-1*96500*0.839

ΔG° = - 80963 J = -80.9 KJ

The Nerst equation gives us the relation of chemical equilibrium and Electric potential.

E=E°- $\frac{RT}{nF} Ln Q$

Where 'R' is the molar gas constant (8.314 J/mol)

It's known that in the equilibrium E=0, so the Nerst equation, at equilibrium, becomes:

E°= $\frac{RT}{nF} Ln K$

Isolating for 'K' gives:

$K=e^{\frac{nFE^{o} }{RT} }$

This shows that 'K' is a fuction of temperature. Assuming this reaction takes place at room temperature (25 °C):

K= $1.53x10^{14}$

6 0

2 years ago