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3 years ago

7

Bombardier beetles inject boiling hydrogen peroxide, or H2O2, into the victims. It then decomposes into hydrogen and oxygen gas.

How many grams of oxygen can be produced if 0.128 grams of hydrogen peroxide are decomposed?

1 answer:

insens350 [35]3 years ago

8 0

Answer:

0.12 g of O2.

Explanation:

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

H2O2 —> H2 + O2

Next, we shall determine the mass of H2O2 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:

Molar mass of H2O2 = (2×1) + (16×2)

= 2 + 32

= 34 g/mol

Mass of H2O2 from the balanced equation = 1 × 34 = 34 g.

Molar mass of O2 = 16×2 = 32 g/mol

Mass of O2 from the balanced equation = 1 × 32 = 32 g

Summary:

From the balanced equation above,

34 g of H2O2 decomposed to produce 32 g of O2.

Finally, we shall determine the mass of O2 produced from the decomposition of 0.128 g of H2O2. This can be obtained as follow:

From the balanced equation above,

34 g of H2O2 decomposed to produce 32 g of O2.

Therefore, 0.128 g of H2O2 will decompose to produce = (0.128×32)/34 = 0.12 g of O2.

Therefore, 0.12 g of O2 was produced.

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When metallic aluminum is added to a solution containing iron(II) sulfate a reaction occurs. What species is being oxidized in t

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<u>Answer:</u> Aluminium is getting oxidized in the given chemical reaction.

<u>Explanation:</u>

Oxidation reaction is defined as the chemical reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the chemical reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

For the given chemical reaction:

$2Al(s)+3FeSO_4(aq.)\rightarrow Al_2(SO_4)_3(aq.)+3Fe(s)$

The half cell reactions for the above reaction follows:

<u>Oxidation half reaction:</u> $Al\rightarrow Al^{3+}+3e^-$

<u>Reduction half reaction:</u> $Fe^{2+}+2e^-\rightarrow Fe$

As, aluminium is loosing 3 electrons to form aluminium cation. Thus, it is getting oxidized. Iron is gaining 2 electrons to form iron anion. Thus, it is getting reduced.

Hence, the oxidized species of the given reaction is aluminium.

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3 years ago

How many total atoms are in 0.670 g of p2o5?

scoray [572]

The molecular formula for compound is $P_{2}O_{5}$ mass of compound is 0.670 g.

To calculate number of atoms first calculate number of moles in the compound as follows:

$n=\frac{m}{M}$

Molar mass of $P_{2}O_{5}$ is 283.886 g/mol, thus,

$n=\frac{0.670 g}{283.886 g/mol}=0.00236 mol$

Thus, number of mole of $P_{2}O_{5}$ is 0.00236 mol.

From the molecular formula 1 mole of $P_{2}O_{5}$ has 2 mol of P (phosphorus) and 5 mol of O (oxygen).

Thus, number of moles of P and O in 0.00236 mol of $P_{2}O_{5}$ will be:

$n_{P}=2\times 0.00236 mol=0.00472 mol$

Similarly,

$n_{O}=5\times 0.00236 mol=0.0118 mol$

Now, in 1 mol of an element there are $6.023\times 10^{23}$ atoms.

Number of atoms of P will be:

$N_{P}=0.00472\times 6.023\times 10^{23}=2.84\times 10^{21}atoms$

Similarly, number of atoms of O will be:

$N_{O}=0.0118\times 6.023\times 10^{23}=7.11\times 10^{21}atoms$

Total number of atoms will be sum of number of atoms of P and O:

$N_{Total}=N_{P}+N_{O}=2.84\times 10^{21}+7.11\times 10^{21}=9.95\times 10^{21} atoms$

Therefore, total number of atoms in $P_{2}O_{5}$ will be $9.95\times 10^{21} atoms$ .

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3 years ago

(b) The conductivity of a 0.01 mol dm–3 solution of a monobasic organic acid in water is 5.07 × 10–2 S m–1. If the molar conduct

Zarrin [17]

Explanation:

The given data is as follows.

$\Lambda^{o}_{m}$ (NaCl) = $1.264 \times 10^{-2}$

$\Lambda^{o}_{m}$ (H-O=C-ONO) = $1.046 \times 10^{-2}$

$\Lambda^{o}_{m}$ (HCl) = $4.261 \times 10^{-2}$

Conductivity of monobasic acid is $5.07 \times 10^{-2} S m^{-1}$

Concentration = 0.01 $mol/dm^{3}$

Therefore, molar conductivity ( $\Lambda_{m}$ ) of monobasic acid is calculated as follows.

$\Lambda_{m} = \frac{conductivity}{concentration}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol/dm^{3}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{0.01 mol \times 10^{3}}$

= $5.07 \times 10^{-3} S m^{2} mol^{-1}$

Also, $\Lambda^{o}_{m}$ = $\Lambda^{o}_{m}_{(HCl)} + \Lambda^{o}_{m}_{(H-O=C-ONO)} - \Lambda^{o}_{m}_{(NaCl)}$

= $4.261 \times 10^{-2} + 1.046 \times 10^{-2} - 1.264 \times 10^{-2}$

= $4.043 \times 10^{-2} S m^{2} mol^{-1}$

Relation between degree of dissociation and molar conductivity is as follows.

$\alpha = \frac{\Lambda_{m}}{\Lambda^{o}_{m}}$

= $\frac{5.07 \times 10^{-2} S m^{-1}}{4.043 \times 10^{-2} S m^{2} mol^{-1}}$

= 0.1254

Whereas relation between acid dissociation constant and degree of dissociation is as follows.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

Putting the values into the above formula we get the following.

K = $\frac{c \times \alpha^{2}}{1 - \alpha}$

= $\frac{0.01 \times (0.1254)^{2}}{1 - 0.1254}$

= $0.017973 \times 10^{-2}$

= $1.7973 \times 10^{-4}$

Hence, the acid dissociation constant is $1.7973 \times 10^{-4}$ .

Also, relation between $pK_{a}$ and $K_{a}$ is as follows.

$pK_{a} = -log K_{a}$

= $-log (1.7973 \times 10^{-4})$

= 3.7454

Therefore, value of $pK_{a}$ is 3.7454.

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explain how energy changes from one form to another in a exothermic reaction. in an endothermic reaction.

Alexxx [7]

Answer:

Exothermic reaction: In exothermic reaction, energy is transferred to the surroundings, and the surrounding temperature increases, this is known as exothermic reaction. In other words energy exits in exothermic reaction. Some example of exothermic reactions are:

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2) Combustion reaction.

3) Some oxidation reaction.

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1) Thermal decomposition.

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