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Aleks [24]
3 years ago
10

Which substance is most likely metallic

Chemistry
2 answers:
Montano1993 [528]3 years ago
8 0
I don’t know “which substance” are you talking about??

You can put the picture there so we can see.
erastova [34]3 years ago
6 0

Answer:

The answer is C

Explanation: I got it right on edg

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The rate of disappearance of HBr in the gas phase reaction 2HBr(g) → H2 (g) I2 (g) is 0.190 M s-1 at 150°C. The rate of appearan
Sunny_sXe [5.5K]

Answer:

The rate of appearance of hydrogen gas is 0.095 M/s.

Explanation:

Rate of the reaction is the change in concentration of of any one of the reactants or products per unit time.

2HBr(g)\rightarrow H_2 (g) I_2 (g)

Given:

The rate of disappearance of HBr = -\frac{d[HBr]}{dt}=0.190 M/s

Rate of the reaction is given by:

R=-\frac{1}{2}\frac{d[HBr]}{dt}=\frac{1}{1}\frac{d[H_2]}{dt}

R=-\frac{1}{2}\frac{d[HBr]}{dt}=\frac{1}{2}\times 0.190 M/s=0.095 M/s

Rate of appearance of the hydrogen gas:

\frac{d[H_2]}{dt}=R =0.095 M/s

The rate of appearance of hydrogen gas is 0.095 M/s.

3 0
3 years ago
Which of the following can be mixed in solution with H₂CO3 to make a buffer?OA. Na₂CO3OB. HFOC. NH3OD. NaHCO3
masha68 [24]

A buffer system usually will need a weak acid, like H2CO3, which is carbonic acid, and a strong base, and the only compound that could figure a fairly strong base among the options given is Na2CO3. NH3 and NaHCO3 are weak bases and HF is a strong acid. Therefore the best answer will be letter A

6 0
1 year 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
Which of these data points is the most reliable?
Sonja [21]

Answer:

third point

Explanation:

omniscient because it gives info about every character instead of one  

hope this helps

5 0
2 years ago
A student dissolves 3.9g of aniline (C6H5NH2) in 200.mL of a solvent with a density of 1.05 g/mL . The student notices that the
tino4ka555 [31]

Answer:

2.1 × 10⁻¹ M

2.0 × 10⁻¹ m

Explanation:

Molarity

The molar mass of aniline (solute) is 93.13 g/mol. The moles corresponding to 3.9 g are:

3.9 g × (1 mol/93.13 g) = 0.042 mol

The volume of the solution is 200 mL (0.200 L). The molarity of aniline is:

M = 0.042 mol/0.200 L = 0.21 M = 2.1 × 10⁻¹ M

Molality

The moles of solute are 0.042 mol.

The density of the solvent is 1.05 g/mL. The mass corresponding to 200 mL is:

200 mL × 1.05 g/mL = 210 g = 0.210 kg

The molality of aniline is:

m = 0.042 mol/0.210 kg = 0.20 m = 2.0 × 10⁻¹ m

5 0
3 years ago
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