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sveticcg [70]
3 years ago
11

The empirical formula of styrene is CH; the molar mass of styrene is 104.14 g/mol. How many H atoms are present in a 7.80 g samp

le of styrene?
Chemistry
1 answer:
Makovka662 [10]3 years ago
3 0

Answer:

In 7.80 g of styrene, we have 3.60×10²³ atoms of H

Explanation:

Empirical formula of styrene is CH

Molecular formula of styrene is C₈H₈

So, 1 mol of styrene has 8 moles of C and 8 moles of H and 1 mol weighs 104.14 grams. Let's make a rule of three:

104.14 g (1 mol of C₈H₈) have 8 moles of H

Then 7.80 g would have ( 7.80  .8) / 104.14 = 0.599 moles

As we know, 1 mol of anything has NA particles (Avogadro's Number,  6.02×10²³), so 0.599 moles will have (mol . NA) particles

0.599 mol . 6.02×10²³ atoms / 1 mol = 3.60×10²³ atoms

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The pH can be defined as the negative logarithm of the hydrogen ion concentration of the solution.

<h3>What is the pH?</h3>

What we call the pH can be defined as the negative logarithm of the hydrogen ion concentration of the solution. We are aware that we can use the relation [H+]  [OH-]  = 1 * 10^-14 to handle the enormity of this problem.

Now, let us go about solving the problems;

1. pH = -log(1 x 10-7) = 7

2.  [H+]=  1 * 10^-14/ 1 x 10^-3

pH =  -log(  1 * 10^-11)

pH = 11

3. pH = -log( 1 x 10^-2)

pH = 2

4. pH = -log(  1 x 10^-10)

pH = 10

5.  [H+]=  1 * 10^-14/ 1 x 10^-8

[H+]=  1 * 10^-6

pH = 6

Learn more about pH:brainly.com/question/1528974

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6 0
2 years ago
Inorganic compounds contain carbon true or false
Sladkaya [172]
Hello,

The correct answer would be:

True.

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8 0
3 years ago
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SEP Analyze Data Look at the phase diagram. What is the state of matter for methanol at a temperature of −0.1°C and a pressure o
devlian [24]
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8 0
3 years ago
Determine the standard enthalpy of formation in kJ/mol for NO given the following information about the formation of NO2 under s
Zarrin [17]

Answer:

90.3 kJ/mol

Explanation:

Let's consider the following thermochemical equation.

2 NO(g) + O₂(g) → 2 NO₂(g)  ∆H°rxn = –114.2 kJ

We can find the standard enthalpy of formation for NO using the following expression.

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × ΔH°f(O₂(g))

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g)) - 1 mol × 0 kJ/mol

∆H°rxn = 2 mol × ΔH°f(NO₂(g)) - 2 mol × ΔH°f(NO(g))

ΔH°f(NO(g)) = (2 mol × ΔH°f(NO₂(g)) - ∆H°rxn) / 2 mol

ΔH°f(NO(g)) = (2 mol × 33.2 kJ/mol + 114.2 kJ) / 2 mol

ΔH°f(NO(g)) = 90.3 kJ/mol

8 0
3 years ago
What is the final volume?
zaharov [31]

Answer:

Option A. 9.4 L

Explanation:

From the question given above, the following data were obtained:

Initial volume (V₁) = 8 L

Initial temperature (T₁) = 293 K

Final temperature (T₂) = 343 K

Final volume (V₂) =?

V₁ / T₁ = V₂ / T₂

8 / 293 = V₂ / 343

Cross multiply

293 × V₂ = 8 × 343

293 × V₂ = 2744

Divide both side by 293

V₂ = 2744 / 293

V₂ = 9.4 L

Therefore, the final volume of the gas is 9.4 L

6 0
3 years ago
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