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lisabon 2012 [21]
3 years ago
10

Given the reaction:

Chemistry
1 answer:
dezoksy [38]3 years ago
3 0

Answer:

V = 22.34 L

Explanation:

Given data:

Volume of O₂ needed = ?

Temperature and pressure = standard

Number of molecules of water produced = 6.0× 10²³

Solution:

Chemical equation:

2H₂ + O₂       →      2H₂O

Number of moles of water:

1 mole contain 6.022× 10²³ molecules

6.0× 10²³ molecules ×  1 mole  /  6.022× 10²³ molecules

0.99 mole

Now we will compare the moles of oxygen and water.

                  H₂O         :            O₂  

                    2            :              1

               0.996         :          0.996

Volume of oxygen needed:

PV = nRT

V = nRT/P

V = 0.996 mol × 0.0821 atm.L/mol.K ×  273.15 K / 1 atm

V = 22.34 L

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The cation affects the intensity of the color more than the color of the solution.

Explanation:

According to Beer Lambert law, the intensity of the colour of the solution depends on the concentration of the specie responsible for the colour in the solution.

Let us recall that transition metal compounds are coloured in solution due to electronic transitions.

Therefore, the cation affects the intensity of the color more than the color of the solution.

7 0
2 years ago
What mass of solid that has a molar mass of 46.0 g/mol should be added to 150.0 g of benzene to raise the boiling point of benze
enyata [817]

Answer : 17.12 g

Explanation:\Delta T =k_b\times m

\Delta T = elevation in boiling point

k_b = boiling point elevation constant

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given \Delta T=6.28^{\circ}C

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Weight of the solvent = 150.0 g = 0.15 kg

Putting in the values

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3 0
3 years ago
Write the complete balanced equation for the following reaction: Al + Fe(NO2)2 Fe + Al(NO2)3
algol13
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An alpha particle is equivalent to the nucleus of an atom of which element?(1 point)
Monica [59]

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7 0
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The elements from this section of the periodic table all belong to the same
krok68 [10]

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Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

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Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell.

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7 0
2 years ago
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