Why we add PH10 When we determine the zn ion

If the molar heat of combustion of liquid benzene at constant volume and 300k is -3272KJ. Calculate the heat of combustion at co

vladimir2022 [97]

Answer:

The heat at constant pressure is -3,275.7413 kJ

Explanation:

The combustion equation is 2C₆H₆ (l) + 15O₂ (g) → 12CO₂ (g) + 6H₂O (l)

$\Delta n_g$ = (12 - 15)/2 = -3/2

We have;

$\Delta H = \Delta U + \Delta n_g\cdot R\cdot T$

Where R and T are constant, and ΔU is given we can write the relationship as follows;

$H = U + \Delta n_g\cdot R\cdot T$

Where;

H = The heat at constant pressure

U = The heat at constant volume = -3,272 kJ

$\Delta n_g$ = The change in the number of gas molecules per mole

R = The universal gas constant = 8.314 J/(mol·K)

T = The temperature = 300 K

Therefore, we get;

H = -3,272 kJ + (-3/2) mol ×8.314 J/(mol·K) ×300 K) × 1 kJ/(1000 J) = -3,275.7413 kJ

The heat at constant pressure, H = -3,275.7413 kJ.

4 0

2 years ago

1. A student tried to solve the following problem by selecting the tile as shown. What, if anything, did the student do wrong?

slega [8]

Answer:

B. The student chose the correct tile, but needs to flip the tile to make the units cancel

Explanation:

Based on the reaction:

2AgNO₃(aq) + Cu(s) → 2Ag(s) + Cu(NO₃)₂ (aq)

<em>2 moles of AgNO₃ react per mole of Cu producing 2 moles of Ag and 1 mole of Cu(NO₃)₂</em>

Thus, if you want to produce 6.75moles of Cu(NO₃)₂ you need:

$6.75moles of Cu(NO_3)_2 \frac{2 mol AgNO_3 }{1 mol Cu(NO_3)_2}$ = 13.50 moles of AgNO₃ are needed

Thus, if you analize the tile shown by the student:

<em>B. The student chose the correct tile, but needs to flip the tile to make the units cancel</em>

6 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.

6 0

2 years ago

What are two ways that scientists can report results of their experiments

tiny-mole [99]

Answer: Scientists often communicate their research results in three general ways. One is to publish their results in peer-reviewed journals that can be ready by other scientists. Two is to present their results at national and international conferences where other scientists can listen to presentations.

Explanation:

7 0

3 years ago

The range in size of most atomic radii is approximately

dsp73

2-5 cm. I hope it helps

5 0

3 years ago

Why we add PH10 When we determine the zn ion​

Why we add PH10 When we determine the zn ion