All categories

3 years ago

Reaction 1: C3H8(g)+5O2(g)→3CO2(g)+4H2O(g), ΔH1=−2043 kJ

1 answer:

6 0

Given teh equation adn the heat of reaction, reaction 2's heat of reaction can be obtained by simply multiplying teh heat of reaction of 1 by 3. The final answer is -6129 kJ.

You might be interested in

Air at 25°C with a dew point of 10 °C enters a humidifier. The air leaving the unit has an absolute humidity of 0.07 kg of water

madreJ [45]

Explanation:

The given data is as follows.

Temperature of dry bulb of air = $25^{o}C$

Dew point = $10^{o}C$ = (10 + 273) K = 283 K

At the dew point temperature, the first drop of water condenses out of air and then,

Partial pressure of water vapor $(P_{a})$ = vapor pressure of water at a given temperature $(P^{s}_{a})$

Using Antoine's equation we get the following.

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{T(K) - 46.854}$

$ln (P^{s}_{a}) = 16.26205 - \frac{3799.887}{283 - 46.854}$

= 0.17079

$P^{s}_{a}$ = 1.18624 kPa

As total pressure $(P_{t})$ = atmospheric pressure = 760 mm Hg

= 101..325 kPa

The absolute humidity of inlet air = $\frac{P^{s}_{a}}{P_{t} - P^{s}_{a}} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

$\frac{1.18624}{101.325 - 1.18624} \times \frac{18 kg H_{2}O}{29 \text{kg dry air}}$

= 0.00735 kg $H_{2}O$ / kg dry air

Hence, air leaving the humidifier has a has an absolute humidity (%) of 0.07 kg $H_{2}O$ / kg dry air.

Therefore, amount of water evaporated for every 1 kg dry air entering the humidifier is as follows.

0.07 kg - 0.00735 kg

= 0.06265 kg $H_{2}O$ for every 1 kg dry air

Hence, calculate the amount of water evaporated for every 100 kg of dry air as follows.

$0.06265 kg \times 100$

= 6.265 kg

Thus, we can conclude that kg of water the must be evaporated into the air for every 100 kg of dry air entering the unit is 6.265 kg.

3 0

3 years ago

A volume of 10.0L of gas at a temperature of 5c is cooled to a temperature of 85C at constant pressure what is the new volume of

Lubov Fominskaja [6]

I don't know how 5°C cooled to 85°C but the answer would be 12.878L

4 0

3 years ago

Read 2 more answers

Write the neutralization equation that would take place in the stomach with two different bases used in antacid products

Aleksandr [31]

In order for the human being stomach to digest food, it uses hydrochloric acid (HCl). This is a very effective an dstrong acid that can burn the lining of the stomach and even cause ulcer if this lining does not contain enough mucus.

Back on track, a reaction between an acid and a base usually yields a salt and water. The two bases that we will use in the reaction with HCl are Mg(OH)2 and CaCO3 (note that (OH)- and (CO3)- both define basis)
MAKE SURE TO BALANCE YOU EQUATION, WHERE THE NUMBER OF MOLECULES ENTERING SHOULD BE EQUAL TO THE NUMBER OF MOLECULES RESULTING.

Equation #1:
<span>2HCl + Mg(OH)2 --> 2H2O + MgCl2
MgCl2 is the salt
</span>Equation#2:
<span>2HCl + CaCO3 --> H2O + CO2 + CaCl2</span>
CaCl2 is the salt
Carbon dioxide is formed as a result of the additional carbon carbon, the reaction ideally outputs H2CO3 which is a very weak base the breaks into water and CO2

5 0

3 years ago

The smell of hot food moves faster than the smell of cold food. Which property of gas is responsible for this?

zhenek [66]

The answer would be kinetic energy. :)

4 0

3 years ago

Read 2 more answers

If two students in a lab are asked to calculate the density of a block with a mass of 31.0 g; they are not told how many digits

Arada [10]

We can see here that the both measurements are accurate. This is the true because the both values are in the correct number of significant figures.

<h3>What is density?</h3>

The term density has to do with the ratio of the mass to the density of the object. We know that the accuracy of a measurement has a lot to do with the number of significant figures in the measurement. In this case, we are told that two students in a lab are asked to calculate the density of a block with a mass of 31.0 g; they are not told how many digits to round their measurements. Student A finds length to be 2 cm width to be 4 cm and height 8 cm. Student B finds the length to be 2.65 cm, width to be 4.20 cm, and height to be 8.35 cm.

For the first student, the density is obtained as;

Density = mass/ volume

= 31.0 g/ 2 cm * 4 cm * 8 cm

= 0.5 g/cm^3

For the second student;

Density = mass/ volume

= 31.0 g/2.65 cm * 4.20 cm * 8.35 cm

= 0.33 g/cm^3

Learn more about density:brainly.com/question/15164682

#SPJ1

3 0

1 year ago