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2 years ago

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How many atoms are in one molecule of dinitrogen pentoxide

1 answer:

Reptile [31]2 years ago

4 0

Explanation:

As the name suggests, a single molecule of Dinitrogen Pentoxide will have 2 (di) nitrogen atoms and 5 (Pent) oxygen atoms.

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user100 [1]

The dependent variable is plant height.

The <em>dependent variable</em> (plant height) is the <em>property that changes</em> as a result something the scientist does.

The <em>independent variable</em> is the <em>property that the scientist changes</em> systematically (the amount of CO_2) to see its effect.

The <em>number of plants</em> and the <em>types of plants</em> are <em>uncontrolled variables</em>. They may or may not affect the heights of the plants.

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2 years ago

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if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d

Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

$mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g$

We can calculate the mass percent of O in MgO using the following expression.

$\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\% = 40.0 \%$

You can learn more about mass percent here: brainly.com/question/14990953

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1 year ago

What does it mean for energy to be conserved

marysya [2.9K]

Answer: It means that the energy is being stored or preserved to be used at a later time.

Your question was a little vague, so if this isn't the answer you were looking for, just let me know and I'll fix it.

Hope this helps! Good luck! :D

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2 years ago

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

<u>Answer:</u> The value of $K_{eq}$ is $4.84\times 10^{-5}$

<u>Explanation:</u>

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

<u>Initial:</u> 0.0280 0.0120

<u>At eqllm:</u> 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

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2 years ago

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kkurt [141]

2 Corsls provide treatments

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2 years ago

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