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frozen [14]
2 years ago
12

7.00g of Compound X with molecular formula C5H10 are burned in a constant-pressure calorimeter containing 35.00kg of water at 25

°C. The temperature of the water is observed to rise by 2.113°C. (You may assume all the heat released by the reaction is absorbed by the water, and none by the calorimeter itself.) Calculate the standard heat of formation of Compound X at 25°C. Be sure your answer has a unit symbol, if necessary, and round it to the correct number of significant digits. [Could you please show a step by step way to solve this]
Chemistry
1 answer:
horsena [70]2 years ago
6 0

Answer:

The standard heat of formation of Compound X at 25°C is -3095.75 kJ/mol.

Explanation:

Mass of compound X = 7.00 g

Moles of compound X = \frac{7.00 g}{70 g/mol}=0.100 mol

Mass of water in calorimeter ,m= 35.00 kg = 35000 g

Change in temperature of the water in calorimeter = ΔT

ΔT = 2.113°C

Specific heat capacity of water ,c= 4.186 J/g °C

Q =  m × c × ΔT

Q=35000 g\times 4.186 J/g ^oC\times 2.113^oC=309,575.6 J=309.575 kJ

Heat gained by 35 kg of water is equal to the heat released on burning of 0.100 moles of compound X.

Heat of formation of Compound X at 25°C:

\frac{-Q}{\text{moles of compound X}}=\frac{-309.575 }{0.100 mol}

= -3095.75 kJ/mol

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Savatey [412]
<h2>Hello!</h2>

The answer is:

When the pressure that a gas exerts  on a sealed container changes from

22.5 psi to 19.86 psi, the  temperature changes from 110°C to

65.9°C.

<h2>Why?</h2>

To calculate which is the last pressure, we need to use Gay-Lussac's law.

The Gay-Lussac's Law states that when the volume is kept constant, the temperature (absolute temperature) and the pressure are proportional.

The Gay-Lussac's equation states that:

\frac{P_1}{T_1}=\frac{P_2}{T_2}

We are given the following information:

We need to remember that since the temperatures are given in Celsius degrees, we need to convert it to Kelvin (absolute temperature) before use the equation, so:

P_1=22.5Psi\\T_1=110\°C=110\°C+273.15=383.15K\\T_1=65.9\°C=65\°C+273.15=338.15K

Now, calculating we have:

\frac{P_1}{T_1}*(T_2)=P_2\\\\P_2=\frac{P_1}{T_1}*(T_2)=\frac{22.5Psi}{383.15}*338.15=19.86Psi

Hence, the final pressure is equal to 19.86 Psi.

Have a nice day!

8 0
2 years ago
Read 2 more answers
Devon’s laboratory is out of material to make phosphate buffer. He is considering using sulfate to make a buffer instead. The pk
sesenic [268]

Answer:

Is not possible to make a buffer near of 7.

Optimal pH for sulfate‑based buffers is 2.

Explanation:

The dissociations of H₂SO₄ are:

H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10

HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.

The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between <em>-11 and -9 and between 1 and 3</em>, having in mind that pH's<0 are not useful. For that reason, <em>is not possible to make a buffer near of 7.</em>

The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is <em>2.</em>

<em />

I hope it helps!

4 0
2 years ago
What are the 4 most abundant minerals in the continental crust? What percentage do these 4 makeup of the total?
Tanya [424]

Answer:

SiO_2, Al_2O_3, CaO, MgO

Explanation:

The four most abundant minerals in the continental crust are, SiO_2, Al_2O_3, CaO, MgO.

The percentages are

SiO_2=60.6\%

Al_2O_3=15.9\%

CaO=6.4\%

MgO=4.7\%

SiO_2 is called quartz

Al_2O_3 is called corundum

CaO is called calcium oxide or quick lime

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5 0
2 years ago
A gold wire has a diameter of 1.00 mm. What length of this wire contains exactly 1.00 mol of gold? (density of Au = 17.0 g/cm3)
allochka39001 [22]

Answer:

The answer to your question is 7160 cm

Explanation:

Data

diameter = 1 mm

length = ?

amount of gold = 1 mol

density = 17 g/cm³

Process

1.- Get the atomic mass of gold

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then, 197g ------------ 1 mol

2.- Calculate the volume of this wire

density = mass/volume

volume = mass/density

volume = 197/17

volume = 5.7 cm³

3.- Calculate the length of the wire

Volume = πr²h

solve for h

h = volume /πr²

radius = 0.05 cm

substitution

h = 5.7/(3.14 x 0.05²)

h = 5.7 / 0.0025

h = 7159.2 cm ≈ 7160 cm

8 0
2 years ago
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1.50x10^-6 ( 1 / 28.01) (6.022x10^23) = 3.22x10^16 molecules CO


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Totol gas particles = 2.05x10^18 molecules

6 0
2 years ago
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