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LenKa [72]
3 years ago
13

Based on the information provided, what is crude oil?

Chemistry
1 answer:
Elena-2011 [213]3 years ago
6 0

Answer:

Choice C - a mixture

Explanation: I just did the question

You might be interested in
Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.
natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

pH=4.76+log(\frac{0.95}{0.92})

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

pH=4.76+log(\frac{0.99}{0.88})

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

pH=4.76+log(\frac{0.85}{1.02})

pH = 4.76 - 0.079

pH = 4.681

5 0
4 years ago
Please help guys if you know it!
iogann1982 [59]
The second one is correct
4 0
3 years ago
How many total atoms are found in one molecule of alpo4?
Kruka [31]
There is 1 aluminium atom and 4 polonium atoms
5 0
4 years ago
Why is it difficult to hold and keep liquids like water in your hand? Guys please help
Zanzabum

Answer:

Molecules move freely around since they don't have a definite shape. This is the reason water stays liquid and couldn't be gripped.

5 0
3 years ago
Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident
Akimi4 [234]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

<em>(1) </em>¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

<em>(2) </em>N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

<em>-(2) </em>2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

<em>(3)</em> N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

<em>-(2) + (3) </em>2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

<em>2×(1)  </em>H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

<em>-(2) + (3) + 2×(1) </em>2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

<em>NH₄Cl(s) → NH₃(g) + HCl(g)</em>

<em>Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol</em>

I hope it helps!

8 0
3 years ago
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