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Bezzdna [24]
2 years ago
11

Se informa que un arco eléctrico estabilizado con agua llegó la temperatura de 25600 °F. En la escala absoluta, ¿Cuál es la rela

ción entre la temperatura y la de una llama de oxiacetileno (3500 °C)?
Chemistry
1 answer:
nignag [31]2 years ago
7 0

Answer:

Explanation:

Translation:

It is reported that an electric arc stabilized with water reached the temperature of 25600 ° F. On the absolute scale, what is the relationship between the temperature and that of an oxyacetylene flame (3500 ° C)?

Solution:

Temperature of the electric arc = 25600°F

Temperature of the oxyacetylene flame = 3500°C

The absolute temperatures are usually recorded in Kelvin. In order to compare the two given temperatures, a need to convert to Kelvin arises.

From celcius to kelvin we have:

     K = 273.15 + T°C

T°C is the temperature in degree celcius

For the oxyacetylene flame:

     K = 273.15 + 3500 = 3773.15K

From fahrenheit to kelvin, we have:

    K = \frac{5}{9}(T °F – 32) + 273.15

T °F is the temperature in degree fahrenheit

For the electric arc:

    K = \frac{5}{9}(25600 – 32) + 273.15 = 14477.59K

We can see that by expressing the two temperature on the same absolute scale, an electric arc is by far hotter than an oxyacetylene flame.

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As with other ionic compounds, potassium bromate, KBrO3, dissociates into ions when it dissolves in water. If 13.8 g of KBrO3 is
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Answer:

ΔH of dissociation is 38,0 kJ/mol

Explanation:

The dissociation reaction of KBrO₃ is:

<em>KBrO₃ → K⁺ + BrO₃⁻ </em>

This dissolution consume heat that is evidenced with the decrease in water temperature.

The heat consumed is:

q = CΔTm

Where C is specific heat of water (4,186 J/mol°C)

ΔT is the temperature changing (18,0°C - 13,0°C = 5,0°C)

And m is mass of water (150,0 mL ≈ 150,0 g)

Replacing, heat consumed is:

q = 3139,5 J ≡ 3,14 kJ

13,8 g of KBrO₃ are:

13,8 g×(1mol/167g) = 0,0826 moles

Thus, ΔH of dissociation is:

3,14kJ / 0,0826mol = <em>38,0 kJ/mol</em>

<em></em>

I hope it helps!

3 0
3 years ago
If 2,035 cal of heat is added to a 500.0 g sample of water at 35.0°C, what is the final
Lynna [10]

Answer:

39.1 °C

Explanation:

Recall the equation for specific heat:

q=mc \Delta T\\

Where q is the heat, m is the mass, c is the specific heat of the substance (in this case water), and delta T is the change in temperature.

You should know that the specific heat of water is 1 cal/g/C.

Using the information in the question:

2035=500(1)(T-35)\\2035=500T-17500\\500T=19535\\T=39.07

The final temperature is about 39.1 °C.

5 0
3 years ago
What does heat mean in chemistry
Tpy6a [65]

Answer:

Heat is the transfer of energy that results from the difference in temperature between a system and its surroundings. At a molecular level, heat is the transfer of energy that makes use of or stimulates disorderly molecular motion in the surroundings.

Explanation:

8 0
3 years ago
At its critical point, ammonia has a density of 0.235 g cm23. You have a special thick­walled glass tube that has a 10.0­mm outs
salantis [7]

Answer:

\large \boxed{\text{69.3 mg}}

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}

4 0
3 years ago
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