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3 years ago

Use the reaction below for the decomposition of sodium azide

Chemistry

1 answer:

kondor19780726 [428]3 years ago

5 0

Answer:

34.05dm^3 of nitrogen gas

Explanation:

First things first, we need to find the number of moles of Sodium azide. We can do that by using the formula m=nM, mass = no. moles x Molar Mass

Rearrange to solve for no. moles and substituting in the known values and we have:

n = m/M

no. moles = 130.0g / (2x(22.99+3x14.01))

no. moles = 130.0/130.0 (4.s.f.)

no. moles = 1

Now we can use the ratio given in the equation to find the number of moles of Nitrogen that will be made:

1 x 3/2 = 1.5 moles of Nitrogen

Now we use the constant that 1 mole of any gas will always have a volume of 22.7dm^3 at STP.

1.5 x 22.7 = 34.05dm^3 of nitrogen gas.

Hope this helped!

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Answer:

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3 years ago

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Calculate the enthalpy change, ∆H in kJ, for the reaction H2O(s) → H2(g) + 1/2O2(g). Use the following information: : +279.9 kJ

Len [333]

Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

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3 years ago