Warm butter- liquid
Liquid nitrogen- gas
paper- solid
neon gas- gas
ice- solid
Answer:
1
Explanation:
zeros are not counted when they come before a natural number
Answer:
You must add 48.97 mL of water to make the 0.200 M diluted solution.
Explanation:
In chemistry, dilution is the reduction in concentration of a chemical in a solution. In other words, it is the process of reducing the concentration of solute in solution, simply adding more solvent to the solution.
In a dilution, the quantity or mass of the solute is not changed but only that of the solvent. As only solvent is being added, by not increasing the amount of solute the concentration of the solute decreases.
The expression for the dilution calculations is:
Cinitial* Vinitial = Cfinal* Vfinal
In this case:
- Cinitial= 12 M
- Vinitial= 0.830 mL
- Cfinal= 0.200 M
- Vfinal= ?
Replacing:
12 M*0.830 mL= 0.200 M*Vfinal
Solving:

Vfinal= 49.8 mL
Since 0.830 mL is the volume you initially have of HCl, the amount of water you must add is:
49.8 mL - 0.830 mL= 48.97 mL
<u><em>You must add 48.97 mL of water to make the 0.200 M diluted solution.</em></u>
Answer:
See explaination
Explanation:
The Cys3-cys97 and cys21-cys142 disulfides restrict the unfolded state of lysozyme enzyme to a class of more compact structures with a less exposed hydrophobic surface, compared to the unfolded states of reduced/non-crosslinked lysozyme. there are 2 major factors which lead to the stabilization of lysozyme due to disulfide bonds-
1- increase in the loop size due to the formation of disulfide bonds that leads to an increase in the even entropic effect.
2- the region formed should be flexible. the strain energy due to the formation of the disulfide bond is lower.
cys21-cys142 has a higher Tm than the cys3-cys97 because it involves flexible parts of the molecule. 21 and 142 residues are located on opposite sides of the active-site cleft where significant hinge-bending motion is seen. this introduces minimal strain in the protein.
<u>Answer:</u> No crystals of potassium sulfate will be seen at 0°C for the given amount.
<u>Explanation:</u>
We are given:
Mass of potassium nitrate = 47.6 g
Mass of potassium sulfate = 8.4 g
Mass of water = 130. g
Solubility of potassium sulfate in water at 0°C = 7.4 g/100 g
This means that 7.4 grams of potassium sulfate is soluble in 100 grams of water
Applying unitary method:
In 100 grams of water, the amount of potassium sulfate dissolved is 7.4 grams
So, in 130 grams of water, the amount of potassium sulfate dissolved will be 
As, the soluble amount is greater than the given amount of potassium sulfate
This means that, all of potassium sulfate will be dissolved.
Hence, no crystals of potassium sulfate will be seen at 0°C for the given amount.