All categories

3 years ago

If 2.19 mol

Chemistry

1 answer:

never [62]3 years ago

3 0

Answer:

1370 °C

Explanation:

We can use the Ideal Gas Law and solve for T.

pV = nRT

Data:

p = 8.12 atm

V = 36.41 L

n = 2.19

R = 0.082 06 L·atm·K⁻¹mol⁻¹

Calculations

$\begin{array} {rcl}pV & = & nRT\\\text{8.12 atm} \times \text{36.41 L} & = & \rm\text{2.19 mol} \times 0.08206 \text{ L}\cdot\text{atm}\cdot\text{K}^{-1}\text{mol}^{-1} \times T\\295.6&=&0.1797T\text{ K}^{-1}\\T& = &\dfrac{294.9 }{\text{0.1797 K}^{-1}}\\\\ & = & \text{1645 K}\\\end{array}$

T = (1645 - 273.15) °C = 1370 °C

The temperature of the sample is 1370 °C.

You might be interested in

Help PLEASE HELP PLEASE

Gennadij [26K]

Answer:

b. k+

Explanation:

4 0

3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

The descriptions below explain two ways that water is used by plants on a sunny day. I. In a process called transpiration, some

eduard

The correct answer for this question is this one: "<span>In transpiration, because some of its properties change, water undergoes a physical change but keeps its identity. In photosynthesis, because its identity changes, water undergoes a chemical change</span>. "
Hope this helps answer your question and have a nice day ahead.

8 0

3 years ago

Read 2 more answers

Making positive lifestyle choices will __________.

Black_prince [1.1K]

Benefit you and your life in many ways!

3 0

3 years ago

In the biosphere, prey represent a food source for predators. An increase in the population of prey means more food for predator

mote1985 [20]

<h2>Answer:</h2>

An increase in the population of prey means more food for predators, which will<u> increase</u> the predators' population. As a result, many more prey will be hunted, causing the prey population to<u> decrease.</u> In response, the predator population will <u>decrease</u>, resulting in an increase in the prey population.

<h3>Explanation:</h3>

The feedback loop is control of the system containing the prey and predator.

In this loop, the population of predator is directly proportional to prey but the population of prey is inversely proportional to a predators population.

As the population of prey increases, the population size of predator becomes large due to the availability of food. This inversely decreases the population of prey, as the predator do more hunt.

So the decrease in prey population results in a decrease in the predator population.

5 0

3 years ago

Read 2 more answers