Ask question

Ask question

All categories

3 years ago

13

Fires in forests can sometimes be stopped by cutting down trees to make a gap (called a 'fire break'). Which part of the fire tr

iangle does this remove?

1 answer:

Akimi4 [234]3 years ago

8 0

It removes the fuel for the fire

You might be interested in

What is true about asteroids

Igoryamba

Answer:

They are big rocks that fly through space and are made of most commonly chondrite. When they collide, they collide with such force that they create craters on places like the moon.

7 0

2 years ago

A closed, frictionless piston-cylinder contains a gas mixture with the following composition on a mass basis: 40% carbon dioxide

Hitman42 [59]

Answer:

W=-37.6kJ, therefore, work is done on the system.

Explanation:

Hello,

In this case, the first step is to compute the moles of each gas present in the given mixture, by using the total mixture weight the mass compositions and their molar masses:

$n_{CO_2}=0.8kg*0.4*\frac{1kmolCO_2}{44kgCO_2}= 0.00727kmolCO_2\\\\n_{O_2}=0.8kg*0.25*\frac{1kmolO_2}{32kgO_2}=0.00625kmolO_2\\ \\n_{Ne}=0.8kg*0.35*\frac{1kmolNe}{20.2kgNe}=0.0139kmolNe$

Next, the total moles:

$n_T=0.00727kmol+0.00625kmol+0.0139kmol=0.02742kmol$

After that, since the process is isobaric, we can compute the work as:

$W=P(V_2-V_1)$

Therefore, we need to compute both the initial and final volumes which are at 260 °C and 95 °C respectively for the same moles and pressure (isobaric closed system)

$V_1=\frac{n_TRT_1}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(260+273)K}{450kPa}=0.27m^3\\ \\V_2=\frac{n_TRT_2}{P}= \frac{0.02742kmol*8.314\frac{kPa*m^3}{kmol\times K}*(95+273)K}{450kPa}=0.19m^3$

Thereby, the magnitude and direction of work turn out:

$W=450kPa(0.19m^3-0.27m^3)\\\\W=-37.6kJ$

Thus, we conclude that since it is negative, work is done on the system (first law of thermodynamics).

Regards.

7 0

3 years ago

Chemistry help!<br><br>Zoom in to see better!!

inna [77]

Answer:

11.9 g of nitrogen monoxide

Explanation:

We'll begin by calculating the number of mole in 6.75 g of NH₃. This can be obtained as follow:

Mass of NH₃ = 6.75 g

Molar mass of NH₃ = 14 + (3×1)

= 14 + 3

= 17 g/mol

Mole of NH₃ =?

Mole = mass /molar mass

Mole of NH₃ = 6.75 / 17

Mole of NH₃ = 0.397 mole

Next, we shall determine the number of mole of NO produced by the reaction of 0.397 mole of NH₃. This can be obtained as follow:

4NH₃ + 5O₂ —> 4NO + 6H₂O

From the balanced equation above,

4 moles of NH₃ reacted to produce 4 moles of NO.

Therefore, 0.397 mole of NH₃ will also react to produce 0.397 mole of NO.

Finally, we shall determine the mass of 0.397 mole of NO. This can be obtained as follow:

Mole of NO = 0.397 mole

Molar mass of NO = 14 + 16 = 30 g/mol

Mass of NO =?

Mass = mole × molar mass

Mass of NO = 0.397 × 30

Mass of NO = 11.9 g

Thus, the mass of NO produced is 11.9 g

7 0

3 years ago

Complete this neutralization equation:<br> H2SO4 + Al(OH)3

elena-s [515]

Answer:

H2SO4 + Al(OH)3 = Al2(SO4)3 + H2O

Explanation:

4 0

4 years ago

In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to

Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

Explanation:

The measured pH of the solution = 5.153

$C_6H_5OH\rightarrow C_6H_5O^-+H^+$

Initially c

At eq'm c-x x x

The expression of dissociation constant is given as:

$K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}$

Concentration of phenoxide ions and hydrogen ions are equal to x.

$pH=-\log[x]$

$5.153=-\log[x]$

$x=7.03\times 10^{-6} M$

$K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}$

$K_a=9.34\times 10^{-11}$

The dissociation constant of phenol from given information is $9.34\times 10^{-11}$ .

4 0

3 years ago