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kvasek [131]
2 years ago
8

What bond vibrations and stretching frequencies should you observe in the IR spectrum of caffeine?

Chemistry
1 answer:
Margarita [4]2 years ago
3 0
  • The bond vibrations that can be observed in the IR spectrum of caffeine :

i) asymmetric, ii) symmetric, iii) wagging, iv) twisting, v) scissoring, vi) rocking

  • while the observable stretching frequencies are ; 1700, 1300, 2900, 1500 and 3000.

Caffeine is a polyatomic molecule ( contains more than two atoms bonded via covalent bonds )  therefore its atoms can vibrate in three dimensions ( x, y, z ).therefore the bond vibrations that can be observed in the IR spectrum of caffeine are :  asymmetric, symmetric, wagging, twisting, scissoring, and rocking.

The observable stretching frequencies that are in the IR spectrum of caffeine ; C = O is 1700 ,  C - H is 1300,  C -H is 2900,  C -N is 1500, and N -H = 3000.

Hence we can conclude The bond vibrations that can be observed in the IR spectrum of caffeine : i) asymmetric, ii) symmetric, iii) wagging, iv) twisting, v) scissoring, vi) rocking and the observable stretching frequencies are ; 1700, 1300, 2900, 1500 and 3000.

Learn more : brainly.com/question/13184210

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How many grams of iron can be made with 21.5g of Fe2O3
SIZIF [17.4K]

The mass (in grams) of iron, Fe that can be made from 21.5 g of Fe₂O₃ is 15.04 g

We'll begin by writing the balanced equation for the reaction. This is given below:

2Fe₂O₃ -> 4Fe + 3O₂

  • Molar mass of Fe₂O₃ = 159.7 g/mol
  • Mass of Fe₂O₃ from the balanced equation = 2 × 159.7 = 319.4 g
  • Molar mass of Fe = 55.85 g/mol
  • Mass of Fe from the balanced equation = 4 × 55.85 = 223.4 g

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

<h3>How to determine the mass of iron, Fe produced</h3>

From the balanced equation above,

319.4 g of Fe₂O₃ decomposed to produce 223.4 g of Fe

Therefore,

21.5 g of Fe₂O₃ will decompose to produce = (21.5 × 223.4) / 319.4 = 15.04 g of Fe

Thus, 15.04 g of Fe were produced.

Learn more about stoichiometry:

brainly.com/question/9526265

#SPJ1

5 0
1 year ago
if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro
tensa zangetsu [6.8K]
<h3>Answers:</h3>

             1) 2 Units of Ozone

             2)  3 Units of Ozone

              3)  9 Units of Ozone

<h3>Solution:</h3>

1)  From 6 Oxygen Particles;

As given,

                          3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      6 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

                           3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      9 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  3 Units of Ozone

3)  From 27 Oxygen Particles;

As given,

                             3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      27 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  9 Units of Ozone

3 0
3 years ago
Now draw the eight planets as circles, and label each one. The scale doesn’t need to be perfect. You can estimate the planets’ s
Ierofanga [76]

i dont now but thaks you po por the pionts

3 0
2 years ago
Read 2 more answers
How do the given animal reproduce ​
Aliun [14]

Answer:

Which animals????????

3 0
3 years ago
Read 2 more answers
Cuántos mililitros de agua hay en 3 litros de una botella de naranjas y las solucion tiene una concentracion de 20% v/v
sertanlavr [38]

Respuesta:

2400 mL

Explicación:

Paso 1: Información dada

  • Volumen de solución: 3 L (3000 mL)
  • Concentración de naranja: 20 % v/v

Paso 2: Calcular el volumen de naranja

La concentración de naranja es de 20 % v/v, es decir, cada 100 mL de solución hay 20 mL de naranja.

3000 mL Sol × 20 mL Naranja/100 mL Solución = 600 mL Naranja

Paso 3: Calcular el volumn de agua

El volumen de soluciónes igual a la suma de los volúmenes de naranja y agua.

VSolución = VNaranja + VAgua

VAgua = VSolución - VNaranja

VAgua = 3000 mL - 600 mL = 2400 mL

8 0
3 years ago
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