Astronauts<span> must </span>wear<span> spacesuits whenever they leave a spacecraft and are exposed to the environment of </span>space<span>. In </span>space, there is no air to breath and no air pressure.Space<span> is extremely cold and filled with dangerous radiation. Without protection, an</span>astronaut<span> would quickly die in </span>space<span>.</span>
The density of ice does not affect the melting rate. But, adding an object does affect the melt rate. The reason this is is because when there is an object, there is less to melt. Hence, affecting the melting rate.
Answer:
132g/mole
Explanation:
using the formula PV=nRT should be used to solve for the number of moles (n). R is a constant which is 62.3637 L mmHG/mole K.
Inorder for your units to match you will have to convert 125ml to .125L and the temperature of 85C to K . you do that by adding 273 to the 85C and get 358K. Once you solve for n then you use that number and divide by the number of grams from the question (.560g) since molar mass is grams/moles.
Networks of feeding relationships is correct
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100