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Mariana [72]
3 years ago
12

Ammonium perchlorate (NH4ClO4) is the solid rocket fuel used by the U.S. Space Shuttle. It reacts with itself to produce nitroge

n gas (N2), chlorine gas (Cl2), oxygen gas (O2), water (H2O), and a great deal of energy. What mass of nitrogen gas is produced by the reaction of 9 gr of ammonium perchlorate? Round your answer to significant digits.

Chemistry
1 answer:
Temka [501]3 years ago
6 0

Answer:

1.08 grams of nitrogen

Explanation:

Thermally ammonium perchlorate dissociate as shown in figure.

Thus as per equation two moles of ammonium perchlorate will give four moles of water, two moles of oxygen, one mole of chlorine and one mole of nitrogen,

The molar mass of ammonium perchlorate = 117.5g/mol

Two moles of ammonium perchlorate = 2X  117.5 = 234 g

The molar mass of nitrogen = 28g

thus 234 grams of ammonium perchlorate will give 28 grams of nitrogen

Hence 9 grams of ammonium perchlorate will give 1.08 grams of nitrogen

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In 1911, Ernest Rutherford tested the atomic model existing at the time by shooting a beam of alpha particles (42He, helium nucl
STatiana [176]

Answer:

At the time of Rutherford's experiment, the accepted model for the atom was the Thomson plum-pudding model of the atom, in which the atom consists of a "sphere" of positive charge distributed all over the sphere, with tiny negative particles (the electrons) inside this sphere.

In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

- The positive charge of the atom was all concentred in a tiny nucleus

- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

5 0
3 years ago
True or false An organism needs two dominant alleles to express that trait.
vova2212 [387]
No one can do that by him or her self

8 0
3 years ago
How do chemists express the rates of chemical reactions
Inessa [10]

Answer:

They expressed it as rate of change in concentration of reactants or products in a chemical reaction

6 0
3 years ago
Read 2 more answers
Using the equation 2H2+O2=2H2O, when 47g of water are produced, how many grams of hydrogen must react?
Fed [463]

See , from the equation we can see that for forming two mole of H2O 2Mole of H2 has to react.

Mass of 2 Mole H2O is 18*2gm or 36gm.

So for forming 36 gm H2O 2×2 I.e. 4 gm H2 has to take part in reaction.

Therefore, to form 1 gm H2O 4÷36 gm of H2 has to take part.

So, for forming 47gm H2O (4÷36)×47 gm H2 has to take part

I.e. 5.22 gm of H2 has to take part

So, ans is 5.22 gm of hydrogen.

Hope it helps!!!

6 0
3 years ago
Read 2 more answers
Consider the following reaction:
adell [148]

Answer:

1. d[H₂O₂]/dt = -6.6 × 10⁻³ mol·L⁻¹s⁻¹; d[H₂O]/dt = 6.6 × 10⁻³ mol·L⁻¹s⁻¹

2. 0.58 mol

Explanation:

1.Given ΔO₂/Δt…

    2H₂O₂     ⟶      2H₂O     +     O₂

-½d[H₂O₂]/dt = +½d[H₂O]/dt = d[O₂]/dt  

d[H₂O₂]/dt = -2d[O₂]/dt = -2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ = -6.6 × 10⁻³mol·L⁻¹s⁻¹

 d[H₂O]/dt =  2d[O₂]/dt =  2 × 3.3 × 10⁻³ mol·L⁻¹s⁻¹ =  6.6 × 10⁻³mol·L⁻¹s⁻¹

2. Moles of O₂  

(a) Initial moles of H₂O₂

\text{Moles} = \text{1.5 L} \times \dfrac{\text{1.0 mol}}{\text{1 L}} = \text{1.5 mol }

(b) Final moles of H₂O₂

The concentration of H₂O₂ has dropped to 0.22 mol·L⁻¹.

\text{Moles} = \text{1.5 L} \times \dfrac{\text{0.22 mol}}{\text{1 L}} = \text{0.33 mol }

(c) Moles of H₂O₂ reacted

Moles reacted = 1.5 mol - 0.33 mol = 1.17 mol

(d) Moles of O₂ formed

\text{Moles of O}_{2} = \text{1.33 mol H$_{2}$O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{2 mol H$_{2}$O}_{2}} = \textbf{0.58 mol O}_{2}\\\\\text{The amount of oxygen formed is $\large \boxed{\textbf{0.58 mol}}$}

8 0
3 years ago
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