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tiny-mole [99]
3 years ago
5

What type of machine is a CD player?

Chemistry
2 answers:
Liula [17]3 years ago
4 0
It is a compound machine i think, tell me if its right!

xenn [34]3 years ago
3 0

Answer:

C. compound machine.

Explanation:

i just took the test and got it correct!

Please say thanks by liking it, so i know i helped you :)

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What is the molarity of a solution that has .8 mols of solute in .5 L of water?
zloy xaker [14]

Answer:

1.6M

Explanation:

molarity = number of moles/number of Liters = 0.8 mol/0.5 L= 1.6 mol/L or 1.6M

5 0
3 years ago
Read 2 more answers
Calculate the percent saturated fat in the total fat in butter
Talja [164]
About 63% of the fat in butter is saturated fat
7 0
3 years ago
When burning petrol in motor vehicles two gases form which contribute to acid rain. what are the two gases?
sukhopar [10]
Nitrogen and Oxygen react to form Nitrogen Oxides. These gasses irritate the lungs and cause acid rain.
8 0
3 years ago
A galvanic (voltaic) cell consists of an electrode composed of chromium in a 1.0 M chromium(III) ion solution and another electr
svetlana [45]

Answer: The potential of the following electrochemical cell is 1.08 V.

Explanation:

E^0_(Cr^{3+}/Cr)=-0.74V[/tex]

E^0_(Cu^{2+}/Cu)=0.34V[/tex]

The element  with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.

Here Cr undergoes oxidation by loss of electrons, thus act as anode. copper undergoes reduction by gain of electrons and thus act as cathode.

2Cr+3Cu^{2+}\rightarrow 2Cr^{3+}+3Cu

E^0=E^0_{cathode}- E^0_{anode}

Where both E^0 are standard reduction potentials, when concentration is 1M.

E^0=E^0_{[Cu^{2+}/Ni]}- E^0_{[Cr^{3+}/Cr]}

E^0=0.34-(-0.74)=1.08V

Thus the potential of the following electrochemical cell is 1.08 V.

6 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:

5 0
3 years ago
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