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2 years ago

Write a balanced chemical equation depicting the formation of one mole of H2O2(g) from its elements in their standard states.

Chemistry

1 answer:

I am Lyosha [343]2 years ago

7 0

Answer:

$H_2(g)+O_2(g)\rightarrow H_2O_2(g)$

Explanation:

Hello!

In this case, since the formation reaction of a compound is undergone when the pure elements composing it are combined, for gaseous hydrogen peroxide, gaseous diatomic hydrogen and oxygen (standard state) must be combined in order to obtain the gaseous hydrogen peroxide as shown below:

$H_2(g)+O_2(g)\rightarrow H_2O_2(g)$

Whereas it is proved there are two hydrogen and oxygen atoms at each side of the chemical equation and therefore it is balanced.

Best regards!

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The decomposition of AB given here in this balanced equation 2AB (g)⟶ A2 (g) + B2 (g), has rate constants of 8.58 x 10-9 L/mol s

denis-greek [22]

Answer:

3.24 × 10^5 J/mol

Explanation:

The activation energy of this reaction can be calculated using the equation:

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

Where; Ea = the activation energy (J/mol)

R = the ideal gas constant = 8.3145 J/Kmol

T1 and T2 = absolute temperatures (K)

k1 and k2 = the reaction rate constants at respective temperature

First, we need to convert the temperatures in °C to K

T(K) = T(°C) + 273.15

T1 = 325°C + 273.15

T1 = 598.15K

T2 = 407°C + 273.15

T2 = 680.15K

Since, k1= 8.58 x 10-9 L/mol, k2= 2.16 x 10-5 L/mol, R= 8.3145 J/Kmol, we can now find Ea

ln(k2/k1) = Ea/R x (1/T1 - 1/T2)

ln(2.16 x 10-5/8.58 x 10-9) = Ea/8.3145 × (1/598.15 - 1/680.15)

ln(2517.4) = Ea/8.3145 × 2.01 × 10^-4

7.831 = Ea(2.417 × 10^-5)

Ea = 3.24 × 10^5 J/mol

8 0

2 years ago

A sample of gasoline contains various hydrocarbons, which comprise atoms of carbon, hydrogen, and oxygen. The hydrocarbons mix t

Naddika [18.5K]

Answer:

mixture
homogenous mixture (of hydrocarbons)
compound

Explanation:

Mixture can be easily separated by physical methods. Homogeneity and heterogeneity of a mixture is determined by whether the components there in are in a single phase and evenly distributed or not.

A solution has a solute evenly dissolved in solvent to form a liquid substance.

An element is the basic form of substance which cannot be broke down into any other simpler unit.

I hope this was helpful.

4 0

2 years ago

Determine the poh of a 0.348 m ba(oh)2 solution at 25°c.

vladimir1956 [14]

Given:

Concentration of Ba(OH)2 = 0.348 M

To determine:

pOH of the above solution

Explanation:

Based on the stoichiometry-

1 mole of Ba(OH)2 is composed of 1 mole of Ba2+ ion and 2 moles of OH- ion

Therefore, concentration of OH- ion = 2*0.348 = 0.696 M

pOH = -log[OH-] = - log[0.696] = 0.157

Ans: pOH of 0.348M Ba(OH)2 is 0.157

6 0

2 years ago

Read 2 more answers

Student Z completes his/her calibration step for part 3 and finds an average massA to be 95.237 g ± 0.005 g. However, in his/her

Troyanec [42]

It is going to be too low because the mass mistakenly used is lower than the initial.

7 0

2 years ago

Read 2 more answers

A tank of 0.1m3 volume contains air at 25∘C and 101.33 kPa. The tank is connected to a compressed-air line which supplies air at

Dmitriy789 [7]

Answer:

Amount of Energy = 23,467.9278J

Explanation:

Given

Cv = 5/2R

Cp = 7/2R wjere R = Boltzmann constant = 8.314

The energy balance in the tank is given as

∆U = Q + W

According to the first law of thermodynamics

In the question, it can be observed that the volume of the reactor is unaltered

So, dV = W = 0.

The Internal energy to keep the tank's constant temperature is given as

∆U = Cv((45°C) - (25°C))

∆U = Cv((45 + 273) - (25 + 273))

∆U = Cv(20)

∆U = 5/2 * 8.314 * 20

∆U = 415.7 J/mol

Before calculating the heat loss of the tank, we must first calculate the amount of moles of gas that entered the tank where P1 = 101.33 kPa

The Initial mole is calculated as

(P * V)/(R * T)

Where P = P1 = 101.33kPa = 101330Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

So, n = (101330 * 0.1)/(8.314*298)

n = 4.089891232222

n = 4.089

Then we Calculate the final moles at P2 = 1500kPa = 1500000Pa

V = Volume of Tank = 0.1m³

R = 8.314J/molK

T = Initial Temperature = 25 + 273 = 298K

n = (1500000 * 0.1)/(8.314*298)

n = 60.54314465936812

n = 60.543

So, tue moles that entered the tank is ∆n

∆n = 60.543 - 4.089

∆n = 56.454

Amount of Energy is then calculated as:(∆n)(U)

Q = 415.7 * 56.454

Q = 23,467.9278J

3 0

2 years ago