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2 years ago

PLEASE HELPP!!

Chemistry

1 answer:

Ivanshal [37]2 years ago

5 0

Answer:

Answer: The correct answer is Option B.

Explanation:

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}Number of moles=Molar massGiven mass ....(1)

For N_2N2 :

Given mass of nitrogen gas = 10 g

Molar mass of nitrogen gas = 28 g/mol

Putting values in above equation, we get:

\text{Moles of iron oxide}=\frac{10g}{28g/mol}=0.357molMoles of iron oxide=28g/mol10g=0.357mol

The given chemical reaction follows:

N_2+O_2\rightarrow 2NON2+O2→2NO

As, oxygen gas is present in excess. Thus, it is considered as an excess reagent and nitrogen is considered as a limiting reagent because it limits the formation of products.

By Stoichiometry of the reaction:

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Alenkinab [10]

Answer:

greater than 115 N upwards

Explanation:

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3 years ago

The hydrogen chloride (HCl) molecule has an internuclear separation of 127 pm (picometers). Assume the atomic isotopes that make

natta225 [31]

Answer:

the energy of the third excited rotational state $\mathbf{E_3 = 16.041 \ meV}$

Explanation:

Given that :

hydrogen chloride (HCl) molecule has an intermolecular separation of 127 pm

Assume the atomic isotopes that make up the molecule are hydrogen-1 (protium) and chlorine-35.

Thus; the reduced mass μ = $\dfrac{m_1 \times m_2}{m_1 + m_2}$

μ = $\dfrac{1 \times 35}{1 + 35}$

μ = $\dfrac{35}{36}$

∵ 1 μ = 1.66 × 10⁻²⁷ kg

μ = $\\ \\ \dfrac{35}{36} \times 1.66 \times 10^{-27} \ \ kg$

μ = 1.6139 × 10⁻²⁷ kg

$r_o = 127 \ pm = 127*10^{-12} \ m$

The rotational level Energy can be expressed by the equation:

$E_J = \dfrac{h^2}{8 \pi^2 I } \times J ( J +1)$

where ;

J = 3 ( i.e third excited state) &

$I = \mu r^2_o$

$E_J= \dfrac{h^2}{8 \pi \mu r^ 2 \mur_o } \times J ( J +1)$

$E_3 = \dfrac{(6.63 \times 10^{-34})^2}{8 \times \pi ^2 \times 1.6139 \times 10^{-27} \times( 127 \times 10^{-12}) ^ 2 } \times 3 ( 3 +1)$

$E_3= 2.5665 \times 10^{-21} \ J$

We know that :

1 J = $\dfrac{1}{1.6 \times 10^{-19}}eV$

$E_3= \dfrac{2.5665 \times 10^{-21} }{1.6 \times 10^{-19}}eV$

$E_3 = 16.041 \times 10 ^{-3} \ eV$

$\mathbf{E_3 = 16.041 \ meV}$

8 0

3 years ago

What volume of 0.350 m koh is required to react completely with 24.0 ml of 0.650 m h3po4?

BartSMP [9]

The complete balanced chemical equation for this is:

First we calculate the number of moles of H3PO4:

moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol

From stoichiometry, 3 moles of KOH is required for every mole of H3PO4, therefore:

moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole H3PO4) = 0.0468 mol

Calculating for volume given molarity of 0.350 M KOH:

Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7 mL

Answer:

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3 years ago

How many electrons does it take to fill the outer energy level of most atoms

Orlov [11]

Depends on the element it can by up to 3, 8, or maybe 16.

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3 years ago

An earthquake's____<br> located directly above its_____

Kipish [7]

Answer:

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located directly above its <u>The epic Center </u>

Explanation:

<h2>Hope this helps you !! </h2>

6 0

2 years ago