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dedylja [7]
2 years ago
10

Chemistry help!!

Chemistry
1 answer:
Scrat [10]2 years ago
8 0

The student originally has 252 grams of water in this experiment.

LAW OF CONSERVATION OF MASS:

  • The law of conservation of mass explains that matter (mass) can neither be created nor destroyed but can be changed from one form to another.

  • This means that in a chemical reaction, the sum of the masses of the reactants must equate to the total mass of product(s).

  • According to this question, a student conducts an experiment to separate water into hydrogen and oxygen. The student collects 28.0 g of hydrogen and 224.0 g of oxygen.

  • Since hydrogen and oxygen are the constituent elements of water, the sum of their masses must equate the mass of water.

  • Therefore, 224g of oxygen + 28g of hydrogen = 252g of water.

Learn more at: brainly.com/question/24996173?referrer=searchResults

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Reika [66]
Can't understand this. English Please!  :) :)
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3 years ago
9. Ibuprofen contain which of the following two functional groups: (1 point)
stira [4]

Answer:

A and C

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4 0
3 years ago
These questions are from an experiment where we had a mixture of Ferrocene, acetylferrocene and diacetyl ferrocene and we separa
PilotLPTM [1.2K]

Answer:

Explanation:

The polarity of the 3 compounds would be in the order of

Ferrocene < Acetylferrocene < Diacetylferrocene

Your TLC data has to also support this observation . This can be checked by measuring the values of Rf ( Retention factor = distance travelled by solute/solvent ) .The Rf values also has to follow this particular order: -

Ferrocene > acetylferrocene > diacetylferrocene

2) Hexane happens to be a non-polar solvent. The polarity of hexane can be increased if some polar solvents for example, ethyl and methylene chloride etc are added

Therefore, in the increasing order of solvents polarity, we have

Hexane < 1:1 mixture of hexane: methylene chloride < 9:1 mixture of methylene chloride:

3) Chromatographic techniques all have a stationary phase in addition to a mobile phase. In the case of column chromatography, the silica gel will be the stationary phase and the solvent that will be poured will be the mobile phase.

4) The TLC and column chromatography both happen to have the same stationary phase which is the silica gel. Also, the same solvent mixture is used in both the techniques. This makes the result of the 2 to be almost the same. The difference seen between them is that, TLC works against the gravity while on the other hand column chromatography works in the direction of the gravity.

5) The key feature in the IR spectra of the acetylferrocene that will be absent in the spectra of ferrocene is the presence of carbonyl stretching frequency at close to 1700 per cm(cm-1). This peak is easily differentiated between both acetyl ferrocene and ferrocene.

4 0
3 years ago
Calculate the new volume of 1.23 mL of a gas at 32 C is subjected to drop in temperature of 20 degrees Celsius
kotykmax [81]

Answer:

1,15mL = V₂

Explanation:

Based on Charle's law the volume is directely proportional to the absolute temperature in a gas under constant pressure. The equation is:

V₁T₂ = V₂T₁

<em>Where V is volume and T absolute temperature of a gas where 1 is initial state and 2, final state.</em>

The V₁ is 1.23mL

T₁ = 32°C + 273.15 = 305.15K

T₂ = T₁ - 20°C = 285.15K

Replacing:

1.23mL*285.15K = V₂*305.15K

<h3>1,15mL = V₂</h3>

<em />

8 0
2 years ago
Un móvil se mueve con movimiento acelerado. En los segundo 2 y 3 los espacios recorridos son 90 y 120 m, Calcula la velocidad in
faust18 [17]

Answer:

La velocidad inicial es 55 \frac{m}{s}y su aceleración es -10 \frac{m}{s^{2} }

Explanation:

Un movimiento es rectilíneo uniformemente variado, cuando la trayectoria del móvil es una línea recta y su velocidad  varia la misma cantidad en cada unidad de tiempo . Dicho de otra manera, este movimiento se caracteriza por una trayectoria que es una línea recta y la velocidad cambia su módulo de manera uniforme: aumenta o disminuye en la misma cantidad por cada unidad de tiempo. Y la aceleración es constante y no nula (diferente de cero).

En este caso la posición del objeto esta dada por la expresión:

x=x0+v0*t+\frac{1}{2} *a*t^{2}

donde x es la posición del cuerpo en un instante dado, x0 la posición en el instante inicial, v0 la velocidad inicial y a la aceleración.

En este caso, por un lado podes considerar:

  • x= 90 m
  • x0= 0 m
  • v0= ?
  • t= 2
  • a= ?

Reemplazando obtenes:

90=v0*2+\frac{1}{2} *a*2^{2}

90=v0*2+\frac{1}{2} *a*4

90=v0*2+2*a

Y por otro lado tenes:

  • x= 120 m
  • x0= 0
  • v0= ?
  • t= 3
  • a= ?

Reemplazando obtenes:

120=v0*3+\frac{1}{2} *a*3^{2}

120=v0*3+\frac{1}{2} *a*9

120=v0*3+\frac{9}{2} *a

Por lo que tenes el siguiente sistema de ecuaciones:

\left \{ {{2*v0+2*a=90} \atop {3*v0+\frac{9}{2} *a=120}} \right.

Resolviendo por el método de sustitución, que consiste en aislar en una ecuación una de las dos incógnitas para sustituirla en la otra ecuación, obtenes:

Despejando v0 de la primera ecuación:

v0= \frac{90-2*a}{2}

Reemplazando en la segunda ecuación:

120=\frac{90-2*a}{2} *3+\frac{9}{2} *a

Resolviendo:

120=(90-2*a)*\frac{3}{2} +\frac{9}{2} *a

120=135-3*a +\frac{9}{2} *a

120-135=-3*a +\frac{9}{2} *a

-15=\frac{3}{2} *a

\frac{-15}{\frac{3}{2} } =a

-10=a

Reemplazando el valor de a en la expresión despejada anteriormente obtenes:

v0= \frac{90-2*(-10)}{2}

Resolviendo:

v0= \frac{90+20}{2}

v0= \frac{110}{2}

v0=55

<u><em>La velocidad inicial es 55 </em></u>\frac{m}{s}<u><em>y su aceleración es -10 </em></u>\frac{m}{s^{2} }<u><em></em></u>

3 0
3 years ago
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