The temperature of a gas is 30 degrees Celsius and its pressure is 760 torr. If the temperature originally

1 answer:

6 0

T₁ = 40°C + 273.15 = 313.15 Kelvin T₂ = 30°C + 273.15 = 303.15 Kelvin

Solving Gay-Lussac's Law for P₁ we get:
P₁ = P₂ • T₁ ÷ T₂ P₁ = 760 torr • 313.15 K ÷ 303.15 K P₁ = 785.07 torr

Using the calculator, we click on the P1 button.
We then enter the 3 numbers 760 313.15 and 303.15 into the correct boxes then click "CALCULATE" and get our answer of 785.07 torr.

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A student balances the following redox reaction using half-reactions.

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Answer: 6.

Explanation:

1) Aluminum

$Al^0-3e^----\ \textgreater \ Al^{3+}$

So each atom of aluminum lost 3 electrons to pass from 0 oxidation state to 3+ oxidation state.

2) Manganesium

$Mn^{2+}+2e^{-}---\ \textgreater \ Mn$

So, each ion of Mn(2+) gained 2 electrons pass from 2+ oxidation state to 0.

3) Balance

Multiply aluminum half-reaction (oxidation) by 2 and multiply manganesium half-raction (reduction) by 3:

$2Al^{0}-6e^{-}---\ \textgreater \ 2Al^{3+}

3Mn^{2+}+6e^{-}---\ \textgreater \ 3Mn^{0}$

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Add the two half-equations:

$2Al^{0}+3Mn^{2+}----\ \textgreater \ 2Al^{3+}+3Mn^{0}$

As you see the left side has 2 Al, 3Mn, and 3*2 positive charges.

The right side has 2 Al, 3 Mn, and 2*3 positive charges.

So, the equation is balanced.

5) Count the number of electrons involved.

As you see 2 atoms of aluminum lost 6 electrons (3 each).

That is the answer to the question. 6 electrons will be lost.

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Answer:

Ionic character

A. PF₃ > PBr₃ > PCl₃

B. BF₃ > CF₄ > NF₃

C. TeF₄ > BrF₃ > SeF₄

Explanation:

The most electronegative element is fluorine, followed chlorine, phosphorous nitrogen etc.

Atoms with high electronegativity tend to form negative ions.

Ionic compounds formed between elements with high electronegativity difference.

% ionic character is directly proportional to electronegativity difference.
According to Pauling Scale E.n for F(4.0), O(3.5), N(3.0), C(2.5), B(2.0), P(2.19), Se(2.55) , Te (2.1), Cl(3.16) and Br(2.96)
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ΔE.N (Electronegativity difference) between( N and F = 4 - 3 = 1), (B and F = 4 - 2 = 2) , (C and F = 4 - 2.5 = 1.5 )

ΔE.N (Electronegativity difference) between( Se and F = 4 - 2.55 = 1.45), (F and Te = 4 - 2.1 = 1.9) , (F and Br = 4 - 2.19 = 1.81 )

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