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2 years ago

Permanent safety markings are NOT required on which of the following areas on household refrigerators using HC refrigerants?

Chemistry

1 answer:

MrRissso [65]2 years ago

8 0

On the inside of the refrigerator door are NOT required Permanent safety markings

Explanation:

Normally in red color the Permanent safety markings arre done. That are wanted on household refrigerators that accept any refrigerant with an A3 flammability grade such as HCs, on or near any evaporators, any exposed refrigerant tubing, near the machine compartment, and on the exterior of the refrigerator.

Each of these markings shall be in words no short than 6.4 mm (1/4 inch) long. Room residents should remove the place quickly regarding the unexpected release of this refrigerant. "DANGER—Risk of Fire or Explosion. Flammable Refrigerant Used. To Be Repaired Only By Trained Service Personnel. Do Not Puncture Refrigerant Tubing.” is the quote to be marked in the above specified places.

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Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 initial partial pressure

-x -x +x +x react

105.1-x 7.96-x x x equilibrium

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

Read 2 more answers

What element speed of sound is 323 m/s?

Pepsi [2]

Answer:

Speed of Sound

Explanation:

Speed of sound, fluid phases

m/s

notes

WEL 206

use 323 27 °C

CRC 323 27 °C

WEL 319

3 0

2 years ago

consideras util conocer las propiedades extensivas e intensivas de los insumos utilizados para la elaboración de producto ¿por q

Brums [2.3K]

Answer:

Explanation:

No.

Las propiedades físicas de los materiales y sistemas a menudo se pueden clasificar como intensivas o extensivas, según cómo cambia la propiedad cuando cambia el tamaño (o extensión) del sistema. Según la IUPAC, una cantidad intensiva es aquella cuya magnitud es independiente del tamaño del sistema, mientras que una cantidad extensiva es aquella cuya magnitud es aditiva para los subsistemas. Esto refleja las ideas matemáticas correspondientes de media y medida, respectivamente.

Una propiedad intensiva es una propiedad a granel, lo que significa que es una propiedad física local de un sistema que no depende del tamaño del sistema o de la cantidad de material en el sistema. Los ejemplos de propiedades intensivas incluyen temperatura, T; índice de refracción, n; densidad, ρ; y dureza de un objeto.

Por el contrario, propiedades extensivas como la masa, el volumen y la entropía de los sistemas son aditivas para los subsistemas porque aumentan y disminuyen a medida que crecen y se reducen, respectivamente.

Estas dos categorías no son exhaustivas, ya que algunas propiedades, físicas no son exclusivamente intensivas ni extensivas. Por ejemplo, la impedancia eléctrica de dos subsistemas es aditiva cuando, y solo cuando, se combinan en serie; mientras que si se combinan en paralelo, la impedancia resultante es menor que la de cualquiera de los subsistemas.

¡Espero haberte ayudado! :)

7 0

1 year ago

A 15.0 mL sample of 0.013 M HNO3 is titrated with 0.017 M CH$NH2 which he Kb=3.9 X 10-10. Determine the pH at these points: At t

kramer

Answer: The pH of the solution in the beginning is 1.89 and the pH of the solution after the addition of base is

Explanation:

For 1: At the beginning

To calculate the pH of the solution, we use the equation:

$pH=-\log[H^+]$

We are given:

Nitric acid is a monoprotic acid and it dissociates 1 mole of hydrogen ions. So, the concentration of hydrogen ions is 0.013 M

$[H^+]=0.013M$

Putting values in above equation, we get:

$pH=-\log(0.013)\\\\pH=1.89$

For 2:

To calculate the number of moles, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$

For nitric acid:

Molarity of nitric acid solution = 0.013 M

Volume of solution = 15 mL

Putting values in above equation, we get:

$0.013M=\frac{\text{Moles of }HNO_3\times 1000}{15}\\\\\text{Moles of }HNO_3=1.95\times 10^{-4}mol$

For methylamine:

Molarity of methylamine solution = 0.017 M

Volume of solution = 10 mL

Putting values in above equation, we get:

$0.017M=\frac{\text{Moles of }CH_3NH_2\times 1000}{10}\\\\\text{Moles of }CH_3NH_2=1.7\times 10^{-4}mol$

The chemical equation for the reaction of nitric acid and methylamine follows:

$HNO_3+CH_3NH_2\rightarrow CH_3NH_3^++NO_3^-$

As, the mole ratio of nitric acid and methyl amine is 1 : 1. So, the limiting reagent will be the reactant whose number of moles are less, which is methyl amine.

By Stoichiometry of the reaction:

1 mole of methyl amine produces 1 mole of $CH_3NH_3^+$

So, $1.7\times 10^{-4}mol$ of methyl amine will produce = $\frac{1}{1}\times 1.7\times 10^{-4}=1.7\times 10^{-4}\text{ moles of }CH_3NH_3^+$

To calculate the $pK_b$ of base, we use the equation:

$pK_b=-\log(K_b)$

where,

$K_b$ = base dissociation constant = $3.9\times 10^{-10}$

Putting values in above equation, we get:

$pK_b=-\log(3.9\time 10^{-10})\\\\pK_b=9.41$

To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[CH_3NH_3^+]}{[CH_3NH_2]})$

We are given:

$pK_b=9.41$

$[CH_3NH_3^+]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

$[CH_3NH_2]=\frac{1.7\times 10^{-4}}{10+15}=6.8\times 10^{-6}M$

Putting values in above equation, we get:

$pOH=9.41+\log(\frac{6.8\times 10^{-6}}{6.8\times 10^{-6}})\\\\pOH=9.41$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH=14-9.41=4.59$

Hence, the pH of the solution is 4.59

4 0

2 years ago

To convert m-ethylaniline to m-ethylfluorobenzene, it should be treated with nitrous acid followed by _____________.

bezimeni [28]

Answer:

$NaBF_4$

Explanation:

In this reaction, we must exchange the amino group ( $NH_2$ ) for a fluorine atom ( $F$ ). Also, the first step in this reaction is the addition of nitrous acid.

We must remember that the amino group in the presence of nitrous acid produces a diazonium salt. The $N_2$ group is a very good leaving group and many benzene derivatives can be produced from this intermediate (see figure 1).

If what we want is to bond a fluorine atom we must use $NaBF_4$ to be able to produce m-ethylfluorobenzene (see figure 2).

I hope it helps!

6 0

2 years ago