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4 years ago

___ H2O2 → _ H2O + ___ O2

Chemistry

1 answer:

never [62]4 years ago

6 0

Answer : The number of moles of oxygen formed are, 0.255 moles.

Explanation :

The given chemical reaction is:

$H_2O_2\rightarrow H_2O+O_2$

This reaction is an unbalanced chemical reaction because in this reaction number of oxygen atoms are not balanced.

In order to balance the chemical equation, the coefficient '2' put before the $H_2O_2$ and $H_2O$ then we get the balanced chemical equation.

The balanced chemical reaction will be:

$2H_2O_2\rightarrow 2H_2O+O_2$

Now we have to calculate the number of moles of oxygen formed.

From the balanced chemical reaction we conclude that,

As, 2 moles of hydrogen peroxide react to give 1 mole of oxygen

So, 0.51 moles of hydrogen peroxide react to give $\frac{0.51}{2}=0.255$ mole of oxygen

Therefore, the number of moles of oxygen formed are, 0.255 moles.

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qwelly [4]

Use the equation q=ncΔT.
q= heat absorbed our released (in this case 1004J)
n= number of moles of sample ( in this case 2.08 mol)
c=molar heat capacity
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3 years ago

What is the gcf of 16 and 72? a. 2 b. 4 c. 16 d. 8

nadya68 [22]

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4 years ago

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Yes.

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4 years ago

What is an acid according to Arrhenius?

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What is an acid according to Arrhenius?

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Fish breathe the dissolved air in water through their gills. Assuming the partial pressures of oxygen and nitrogen in air to be

amid [387]

Answer:

X(O₂) = 0.323

X(N₂) = 0.677

Explanation:

We have the partial pressures of oxygen (O₂) and nitrogen (N₂):

P(O₂) = 0.20 atm

P(N₂) = 0.80 atm

In order to solve the problem, you need the solubilities of each gas in water at 298 K. We can consider 1.3 x 10⁻³ mol/(L atm) for oxygen (O₂) and 6.8 x 10⁻⁴mol/(L atm) for nitrogen (N₂) from the bibliography.

s(O₂) = 1.3 x 10⁻³ mol/(L atm)

s(N₂) = 6.8 x 10⁻⁴mol/(L atm)

So, we calculate the concentration (C) of each gas as the product of its partial pressure (P) and the solubility (s):

C(O₂) = P(O₂) x s(O₂) = 0.20 atm x 1.3 x 10⁻³ mol/(L atm) = 2.6 x 10⁻⁴mol/L

C(N₂) = P(N₂) x s(N₂) = 0.80 atm x 6.8 x 10⁻⁴mol/(L atm) = 5.44 x 10⁻⁴ mol/L

In 1 liter of water, we have the following number of moles (n):

n(O₂) = 2.6 x 10⁻⁴ mol

n(N₂) = 5.44 x 10⁻⁴ mol

Thus, the total number of moles (nt) is calculated as the sum of the number of moles of the gases in the mixture:

nt = n(O₂) + n(N₂) = 2.6 x 10⁻⁴ mol + 5.44 x 10⁻⁴ mol = 8.04 x 10⁻⁴ mol

Finally, the mole fraction of each gas is calculated as the ratio between the number of moles of each gas and the total number of moles:

X(O₂) = n(O₂)/nt = 2.6 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.323

X(N₂) = n(N₂)/nt = 5.44 x 10⁻⁴ mol/(8.04 x 10⁻⁴ mol) = 0.677

5 0

3 years ago

_____ H2O2 → _____ H2O + _____ O2

___ H2O2 → _ H2O + ___ O2