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podryga [215]
3 years ago
6

How many grams of NaCl do you need to prepare 200 ml of a 5M solution? The molecular weight of NaCl is 58.4 g/mol. Smol 0.2L-1mo

l NaCl lib Imol NaClx 8.glol Nacl Mol 58.19 Nac
Chemistry
1 answer:
Butoxors [25]3 years ago
4 0

Answer:

The answer to your question is: 58.4 g of NaCl

Explanation:

Data

Volume = 200 ml = 0.2 l

Concentration = 5M

MW = 58.4 g

mass NaCl = ?

Formula

             Molarity = (# of moles ) / volume

             # of moles = Molarity x volume

             # of moles = 5 x 0.2

             # of moles = 1

              58.4 g ---------------------- 1 mol

              x          ---------------------  1 mol

              x = (1 x 58.4) / 1

             x = 58.4 g of NaCl

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3 years ago
Addition reactions of alkenes are characterized by _________. A) formation of a bond B) addition of two groups across a double b
mr_godi [17]

Answer:

B. ADDITION OF TWO GROUPS ACROSS A DOUBLE BOND

Explanation:

Addition reaction of alkenes involves the conversion of the double bond in alkenes Inyo single bonds by the addition of two groups of atoms or radicals.

During this addition reaction, two substances, an unsaturated compound(e.g. ethane) and an attacking reagent (hydrogen, halogens, hydrogen halides, chlorine and bromine water) combines to form a single new compound without forming any other products. So a saturated product or one in which is an increase in degree of saturation is formed.

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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:

5 0
3 years ago
Someone please answer this...
kupik [55]

Answer:

5446.8 J

Explanation:

From the question given above, the following data were obtained:

Mass (M) = 50 g

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Specific heat capacity (C) = 0.89 J/gºC

Heat (Q) required =?

Next, we shall determine the change in the temperature. This can be obtained as follow:

Initial temperature (T₁) = 70 °C

Final temperature (T₂) = 192.4 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 192.4 – 70

ΔT = 122.4 °C

Finally, we shall determine the heat required to heat up the block of aluminum as follow:

Mass (M) = 50 g

Specific heat capacity (C) = 0.89 J/gºC

Change in temperature (ΔT) = 122.4 °C

Heat (Q) required =?

Q = MCΔT

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Q = 5446.8 J

Thus, the heat required to heat up the block of aluminum is 5446.8 J

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Answer: (C)

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The relation between the frequency and wavelength of a wave is

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The frequency of a wave is number of complete cycles passing a particular point per second. Its S.I unit is Hertz whereas the wavelength of a wave is the distance between two consecutive crest and trough in meters.

So, on increasing the frequency of a wave, there will be more number of the cycles of wave per second which will decrease the distance between the consecutive crest and trough i.e wavelength.

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3 years ago
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