Ask question

Ask question

All categories

3 years ago

13

Why do onions make you cry?

1 answer:

Helga [31]3 years ago

3 0

When you cut an onion, it releases a gas, when mixed with certain enzymes in the onion, it creates a sulfur gas. These gases then get to your eyes and create a mild acid which irritates the eyes. Hope this helps, if you have any questions please feel free to message me ;)

You might be interested in

when your tires heat up from driving on a hot road the pressure inside of them increase. this is an example of which gas law

sergiy2304 [10]

Le Châtelier's Principle

This principle states that as temperature increases the pressure must also increase to remain at equilibrium

6 0

3 years ago

zimovet [89]

Answer:

if its a multiple question answer its hydroden and carbon

if not its carbon

3 0

2 years ago

Help out please I really don’t understand these

gavmur [86]

In English

7. Name the elements found in nature in diatomatic form.

8. What is the difference between double and triple single link?

4 0

2 years ago

How are the sensory clues from watching a fire different from the sensory clues from watching a firefly?

Mademuasel [1]

Answer:

watching a firefly is much different than watching a fire because watching a fire can make you hot and watching a firefly can make you happy

7 0

2 years ago

For the equilibrium

Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

$H_2S=0.596M$

$H_2=0.004 M$

$S_2=0.002 M$

Explanation:

We are given that for the equilibrium

$2H_2S\rightleftharpoons 2H_2(g)+S_2(g)$

$k_c=9.0\times 10^{-8}$

Temperature, $T=700^{\circ}C$

Initial concentration of

$H_2S=0.30M$

$H_2=0.30 M$

$S_2=0.150 M$

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles of reactant reduced and form product

Concentration of

$H_2S=0.30+2x$

$H_2=0.30-2x$

$S_2=0.150-x$

At equilibrium

Equilibrium constant

$K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}$

Substitute the values

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

$9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}$

By solving we get

$x\approx 0.148$

Now, equilibrium concentration of gases

$H_2S=0.30+2(0.148)=0.596M$

$H_2=0.30-2(0.148)=0.004 M$

$S_2=0.150-0.148=0.002 M$

3 0

2 years ago