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Llana [10]
3 years ago
9

If you are using 3.00% (mass/mass) hydrogen peroxide solution and you determine that the mass of solution required to reach the

equivalence point is 5.125 g, how many moles of hydrogen peroxide molecules are present?
Chemistry
1 answer:
Maurinko [17]3 years ago
5 0

Answer:

0.004522 moles of hydrogen peroxide molecules are present.

Explanation:

Mass by mass percentage of hydrogen peroxide solution = w/w% = 3%

Mass of the solution , m= 5.125 g

Mass of the hydrogen peroxide = x

w/w\% = \frac{x}{m}\times 100

3\%=\frac{x}{5.125 g}\times 100

x=\frac{3\times 5.125 g}{100}=0.15375 g

Mass of hydregn pervade in the solution = 0.15375 g

Moles of hydregn pervade in the solution :

=\fraC{ 0.15375 g}{34 g/mol}=0.004522 mol

0.004522 moles of hydrogen peroxide molecules are present.

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what is the oxidation state of each element in coh2? c o h what is the oxidation state of each element in febr3? fe br
fgiga [73]

The oxidation state of the elements  in the compounds are:

CoH₂:

  • Co = +2
  • H = -1

FeBr₃:

  • Fe = +3
  • Br = -1

<h3>What is the oxidation states of the elements in the given compounds?</h3>

The oxidation states of the elements in each of the given compounds is determined as follows:

Cobalt dihydride, CoH₂

Co = +2

H = -1

Iron (iii) bromide, FeBr₃

Fe = +3

Br = -1

In conclusion, the oxidation state of the elements are charges they have in the compound.

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8 0
2 years ago
In order to study hydrogen halide decomposition, a researcher fills an evacuated 1.11 L flask with 0.822 mol of HI gas and allow
Elza [17]

Answer:

Kc = 168.0749

Explanation:

  •           2HI(g)     ↔    H2(g) + I2(g)

initial mol:   0.822               0          0

equil. mol: 2(0.822 - x)         x           x

∴ [ HI ]eq = 0.055 mol/L = 2(0.822 - x) / (1.11 L )

⇒ 1.644 - 2x = 0.055 * 1.11

⇒ 1.644 = 2x + 0.06105

⇒ 2x = 1.583

⇒ x = 0.7915 mol equilibrium

⇒ [ H2 ] eq = 0.7915mol / 1.11L = 0.7130 M = [ I2 ] eq

⇒ Kc = ([ H2 ] * [ I2 ]) / [ HI ]²

⇒ Kc = ( 0.7130² ) / ( 0.055² )

⇒ Kc = 168.0749

 

4 0
3 years ago
What is the kinetic energy of the emitted electrons when cesium is exposed to UV rays of frequency 1.3×1015Hz?
Sonja [21]

Answer:

We know that

ħf = ф + Ekmax  

where  

ħ = planks constant = 6.626x10^-34 J s  

f = frequency of incident light = 1.3x10^15 /s (1 Hz =

1/s)

ф = work function of the cesium = 2.14 eV  

Ekmax = max  kinetic energy of the emmitted electron.  

We distinguish that:

1 eV = 1.602x10^-19 J

So:

2.14 eV x (1.602x10^-19 J / 1 eV) = 3.428x10^-19 J

So,

Ekmax = (6.626x10^-34 J s) x (1.3x10^15 / s) - 3.428x10^-19 J

= 8.6138x10^-19 J - 3.428x10^-19 J = 5.1858x10^-19 J  

Answer:

5.19x10^-19 J  

Kinetic energy:

In physics, the kinetic energy of an object is the energy that it owns due to its motion. It is defined as the work required accelerating a body of a given mass from rest to its specified velocity. Having expanded this energy during its acceleration, the body upholds this kinetic energy lest its speed changes.

Answer details:

Subject: Chemistry

Level: College

Keywords:

• Energy  

• Kinetic energy

• Kinetic energy of emitted electrons

Learn more to evaluate:

brainly.com/question/4997492

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