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vesna_86 [32]
3 years ago
5

When gasses are heated the average velocity (speed) of the gas particles decreases.

Chemistry
1 answer:
Kitty [74]3 years ago
6 0
I think that would be true!
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Question: A boy swimming in a lake smacks the surface of the water with his hand, creating sound waves that travel through the w
rodikova [14]
<span>The speed of sound in air is about 340 m/s, while the speed of sound in water is about 1,500 m/s. In general, sound travels faster in denser media, so the fish hears the noise first.</span>
7 0
3 years ago
You place 36.5 ml of 0.266 M Ba(OH)2 in a coffee-cup calorimeter at 25.00°C and add 56.6 ml of 0.648 M HCl, also at 25.00°C. Aft
olya-2409 [2.1K]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

Explanation :

First we have to calculate the mass of solution.

Mass=Density\times Volume

Volume of solution = Volume of HCl + Volume of Ba(OH)_2

Volume of solution = 56.6 mL + 36.5 mL

Volume of solution = 93.1 mL

Density of solution = 1 g/mL

Mass=1g/mL\times 93.1mL=93.1g

The mass of solution is, 93.1 grams.

Now we have to calculate the heat released in the system.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat released = ?

m = mass = 93.1 g

C_p = specific heat capacity of water = 4.184J/g^oC

T_1 = initial temperature  = 25.0^oC

T_2 = final temperature  = 29.83^oC

Now put all the given value in the above formula, we get:

Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC

Q=1881.43J=1.88kJ        (1 kJ = 1000 J)

Now we have to calculate the moles of Ba(OH)_2 and HCl.

\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}

\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol

and,

\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}

\text{Moles of }HCl=0.648M\times 0.0566L=3.66\times 10^{-2}mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

From the balanced reaction we conclude that

As, 1 mole of Ba(OH)_2 react with 2 mole of HCl

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of HCl

From this we conclude that, HCl is an excess reagent because the given moles are greater than the required moles and Ba(OH)_2 is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2O

From the reaction, we conclude that

As, 1 mole of Ba(OH)_2 react to give 2 mole of H_2O

So, 9.71\times 10^{-3}  moles of Ba(OH)_2 react with 9.71\times 10^{-3}\times 2=0.0194 moles of H_2O

Now we have to calculate the change in enthalpy of the reaction.

\Delta H_{rxn}=-\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 1.88 kJ

n = moles of reaction = 0.0194 mole

Now put all the given values in above expression, we get:

\Delta H_{rxn}=-\frac{1.88kJ}{0.0194mole}=-96.9kJ/mole

The negative sign indicates that the heat is released.

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, -96.9 kJ/mole

3 0
3 years ago
Formula los siguientes compuesto: Dietil eter, Etanol, Propanotriol, Acido Propanodioico, Pentanal, Pentano-2,4-diona, Metanoato
9966 [12]

Answer:

Explanation:

En este caso para formular los compuestos, debes identificar el grupo funcional principal de la molecula. Una vez que eso está hecho, puedes intentar formularlo.

Empezaremos primero identificando el grupo funcional principal de la molécula, para luego formularlo correctamente.

Dietil eter: la terminación eter al final significa que pertenece al grupo de los éteres, el cual tiene como formula general R - O - R.

Etanol: debido a que termina en ol, este grupo pertenece a los alcoholes. Para formularlo solo se dibuja la molecula del etano, junto a un enlace con el grupo OH, como su formula general R - OH.

Propanotriol: igualmente termina en ol, por lo tanto es un alcohol, sin embargo, en este caso, tambien tiene la terminación prefija tri, asi que significa que hay 3 grupos OH en la molecula.

Acido propanodioico: esta es sencilla, porque empieza como acido, y solo hay un grupo funcional que empieza así y son los acidos carboxilicos, es decir, el grupo COOH (R - COOH) que es el carboxilo. Tiene el prefijo di, antes del oico, por lo que son dos carboxilos presentes en la molecula.

Pentanal: el sufijo al, significa que pertenece al grupo de los aldehidos, en este caso, posee el grupo carbonilo H - C = O.

Pentano - 2,4 - diona: la terminación ona significa que pertenece al grupo de las cetonas, (R - CO - R), parecido a los aldehidos, con la diferencia de que tiene grupos alquilos en lugar de un hidrogeno.

Metanoato de metilo: la terminación ato de ilo, pertenece a los esteres, (R - COOR) derivado de los acidos carboxilicos.

De aqui en adelante solo mencionaré los grupos funcionales pues ya se explicó el por que, por sus terminaciones:

Ciclohexano - 1.3 - diol: este pertenece a los alcoholes.

Acido heptanoico: acido carboxilico

Ciclobutil metil eter: eteres

Acetato de etilo: ester

2-metilbenzaldehído: aldehído unido a un grupo aromatico como el benceno.

Ciclohexanona: un ciclo (cadena cerrada) unido a un grupo carbonilo.

Butanona: cetona.

Observa la foto adjunta para que veas la formulación de cada una:

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This element is used to improve the conductivity of germanium.
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