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3 years ago

How many grams of o2 are required to produce 358.5 grams of zno? 2zn + o2 ® 2zno?

1 answer:

6 0

The balanced equation for the reaction is ;
2Zn + O2 —> 2ZnO
The stoichiometry of O2 to ZnO is 1:2
The mass of ZnO formed - 358.5 g
The number of moles formed - 358.5 g / 81.4 g/mol = 4.4 moles
Therefore number of O2 moles reacted = 4.4 moles /2 = 2.2 mol
Mass of O2 reacted = 2.2 mol x 32 g/mol = 70.4 g

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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

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kozerog [31]

Answer:

Energy is released as heat during the reaction.

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3 years ago

If a substance cannot be separated <br> physically it is?

MArishka [77]

Its an element Im pretty sure

3 0

3 years ago

Read 2 more answers

The volume (in L) that would be occupied by 5.00 mols of 02 at STP is

crimeas [40]

Answer : The volume of oxygen at STP is 112.0665 L

Solution : Given,

The number of moles of $O_2$ = 5 moles

At STP, the temperature is 273 K and pressure is 1 atm.

Using ideal gas law equation :

$PV=nRT$

where,

P = pressure of gas

V = volume of gas

n = the number of moles

T = temperature of gas

R = gas constant = 0.0821 L atm/mole K (Given)

By rearranging the above ideal gas law equation, we get

$V=\frac{nRT}{P}$

Now put all the given values in this expression, we get the value of volume.

$V=\frac{(5moles)\times (0.0821Latm/moleK)\times (273K)}{1atm}=112.0665L$

Therefore, the volume of oxygen at STP is 112.0665 L

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3 years ago

Question 12 A chemistry student must write down in her lab notebook the concentration of a solution of sodium thiosulfate. The c

o-na [289]

Weight of sodium thiosulfate = 76.148 - 8.2

= 67.948 g.

Concentration of the solution = 67.948 / 172.7

= 0.393 g / mL. to the nearest thousandth . (answer).

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3 years ago