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3 years ago

Use the formation reactions below such that when added together, they match the balanced equation for the combustion of methane.

Chemistry

1 answer:

muminat3 years ago

7 0

Answer:

ΔH of the reaction is -802.3kJ.

Explanation:

Using Hess's law, you can know ΔH of reaction by the sum of ΔH's of half-reactions.

Using the reactions:

<em>(1) </em>Cgraphite(s)+ 2H₂(g) → CH₄(g) ΔH₁ = −74.80kJ

<em>(2) </em>Cgraphite(s)+ O₂(g) → CO₂(g) ΔH₂ = −393.5k J

<em>(3) </em>H₂(g) + 1/2 O₂(g) → H₂O(g) ΔH₃ = −241.80kJ

The sum of (2) - (1) produce:

CH₄(g) + O₂(g) → CO₂(g) + 2H₂(g) ΔH' = -393.5kJ - (-74.80kJ) = -318.7kJ

And the sum of this reaction with 2×(3) produce:

CH₄(g) + 2 O₂(g) → CO₂(g) + 2H₂O(g) And ΔH = -318.7kJ + 2×(-241.80kJ) =

<em>-802.3kJ</em>

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RideAnS [48]

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3 years ago

Calculate the molality of acetone in an aqueous solution with a mole fraction for acetone of 0.241. Answer in units of m.

Anit [1.1K]

Answer: The molality of solution is 17.6 mole/kg

Explanation:

Molality of a solution is defined as the number of moles of solute dissolved per kg of the solvent.

$Molarity=\frac{n}{W_s}$

where,

n = moles of solute

$W_s$ = weight of solvent in kg

moles of acetone (solute) = 0.241

moles of water (solvent )= (1-0.241) = 0.759

mass of water (solvent )= $moles\times {\text {Molar Mass}}=0.759\times 18=13.7g=0.0137kg$

Now put all the given values in the formula of molality, we get

$Molality=\frac{0.241}{0.0137kg}=17.6mole/kg$

Therefore, the molality of solution is 17.6 mole/kg

3 0

3 years ago

Which of these substances has the lowest pH? 0.5 M HBr, pOH = 13.5 0.05 M HCl, pOH = 12.7 0.005 M KOH, pOH = 2.3

vagabundo [1.1K]

<h3><u>Answer;</u></h3>

0.5 M HBr, pOH = 13.5 ; Has the lowest pH

<h3><u>Explanation;</u></h3>

From the question;

pH = -Log [OH]

or pH = 14 - pOH

Therefore;

For 0.5 M HBr

[H+] = 0.5 M

pH = - Log [0.5]

= 0.30

For; pOH = 13.5

pH = 14 - pOH

= 14 -13.5

= 0.5

For; 0.05 M HCl

pH = - log [H+]

[H+] = 0.05

pH = - Log [0.05]

= 1.30

For; pOH = 12.7

pH = 14 -pOH

= 14 -12.7

= 1.30

For; 0.005 M KOH,

pOH = - log [OH]

[OH-] = 0.005

pOH = - Log 0.005

= 2.30

pH = 14 - 2.30

= 11.7

For; pOH = 2.3

pH = 14 -pOH

= 14- 2.3

= 11.7

6 0

3 years ago

Read 2 more answers

What is the molarity of a solution of sulfuric acid if 49.0 mL of it are neutralized by a titration

vaieri [72.5K]

Answer:

THE MOLARITY OF THE ACID IS 0.232M

Explanation:

Molarity is the number of moles of solute per liter of a solution.

In titration, the molarity of the acid and the molarity of the base are related by this equation:

Ca VA / CbVb = na / nb

Ca = concentration of the acid = ?

Va = volume of the acid = 49 mL

Cb = concentration of the base = 0.333M

Vb = volume of the base = 68.4mL

Na = number of mole of acid = 1

Nb = number of mole of base = 2

Equation for the reaction:

H2SO4 + 2NaOH --------> Na2SO4 + 2H2O

Solving for the molarity of the acid, we have :

Ca = CbVbNa / VaNb

Ca = 0.333M * 68.4mL * 1 / 49mL *2

Ca = 22.7772 * 10^-3 / 98 *10^-3

Ca = 0.232M

The molarity of the acid is therefore 0.232M.

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3 years ago

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