How many molecules are in 1.00 g of C10H14N2? 3.71 x 1021 9.76 x 1025 1.02 x 10-26 2.32 x 1022

Chemistry

1 answer:

Reika [66]3 years ago

8 0

Answer:

3.71 x 1021 (First option)

Explanation:

Let's determine the molar mass of the compound.

C₁₀H₁₄N₂ = 12. 10 + 14 .1 + 14 .2 = 162 g/mol

So 162 grams of compound are contained in 1 mol

1 g of C₁₀H₁₄N₂ are contained in ( 1 / 162) = 0.00617 moles

Let's conver the moles into amount of molecules (mol . NA)

0.00617 . 6.02×10²³ = 3.71×10²¹ molecules

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Consider the following reaction at constant pressure. Use the information provided below to determine the value of ΔS at 473 K.

Elis [28]

Answer:

The reaction will be spontaneous

Explanation:

To determine if the reaction will be spontaneous or not at this temperature, we need to calculate the Gibbs's energy using the following formula:

$\Delta G= \Delta H - T * \Delta S$

<u>If the Gibbs's energy is negative, the reaction will be spontaneous, but if it's positive it will not.</u>

Calculating the $\Delta G= -1267 - 473 K* \Delta S$ :

$\Delta G= -1267 - 473 K* \Delta S$

Now, other factor we need to determine is the sign of the S variation. When talking about gases, the more moles you have in your system the more enthropic it is.

In this reaction you go from 7 moles to 8 moles of gas, so you can say that you are going from one enthropy to another higher than the first one. This results in: $\Delta S>0[/tex}Back to this expression: [tex]\Delta G= -1267 - 473 K* \Delta S$

If the variation of S is positive, the Gibbs's energy will be negative always and the reaction will be spontaneous.

4 0

3 years ago

The iodine "clock reaction" involves the following sequence of reactions occurring in a reaction mixture in a single beaker. 1.

Mars2501 [29]

C: 0.012 mol.

<h3>Explanation</h3>

Start with 0.0020 moles of iodate ions ${\text{IO}_{3}}^{-}$ .

How many moles of iodine $\text{I}_2$ will be produced?

${\text{IO}_{3}}^{-}$ converts to $\text{I}_2$ in the first reaction. The coefficient in front of $\text{I}_2$ is three times the coefficient in front of ${\text{IO}_{3}}^{-}$ . In other words, each mole of ${\text{IO}_{3}}^{-}$ will produce three moles of $\text{I}_2$ . 0.0020 moles of ${\text{IO}_{3}}^{-}$ will convert to 0.0060 moles of $\text{I}_2$ .

How many moles of thiosulfate ions ${\text{S}_2\text{O}_3}^{2-}$ are required?

$\text{I}_2$ reacts with ${\text{S}_2\text{O}_3}^{2-}$ in the second reaction. The coefficient in front of $\text{I}_2$ is twice the coefficient in front of ${\text{S}_2\text{O}_3}^{2-}$ . How many moles of ${\text{S}_2\text{O}_3}^{2-}$ does each mole of $\text{I}_2$ consume? Two. 0.0060 moles of $\text{I}_2$ will be produced. As a result, $2 \times 0.0060 = 0.0120$ moles of ${\text{S}_2\text{O}_3}^{2-}$ will be needed.