Answer:
The acid-base reaction produces glycine reduction, and hence the increase of glycine pH.
Explanation:
The glycine is an amino acid with the following chemical formula:
NH₂CH₂COOH
The COOH functional group is what gives the acid properties in the molecule.
Hence, when NaOH is added to glycine an acid-base reaction takes place in which COOH reacts with the NaOH added:
NH₂CH₂COOH + OH⁻ ⇄ NH₂CH₂COO⁻ + H₂O
The glycine concentration starts to shift to its ion form (NH₂CH₂COO⁻) because of the reaction with NaOH, that is why the pH glycine increases when NaOH is added.
Therefore, the acid-base reaction produces glycine reduction, and hence the increase of glycine pH.
I hope it helps you!
Answer:
The copper, because its specific heat is higher, meaning it takes more heat (Joules) per gram to raise the temperature 1 degree Celsius.
Explanation:
We know that to relate solutions of with the factors of molarity and volume, we can use the equation:

**
NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.
So now we can assign values to these variables. Let us say that the 18 M

is the left side of the equation. Then we have:

We can then solve for

:

and

or

We now know that the total amount of volume of the 4.35 M solution will be
210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.
The answer is B. The density, or more precisely, the volumetric mass density, of a substance is its mass per unit volume.
Answer:
Option d. 0.10 m Cr₂(SO₄)₃
Explanation:
Formula for the osmotic pressure is determined as:
π = M . R . T . i
So you have to take account the i (Van't Hoff factor, numbers of ions dissolved)
Urea is an organic compound, so the i value is 1
Zync acetate can be dissociated:
Zn(CH₃COO)₂ → 1Zn²⁺ + 2CH₃COO⁻
In this case, the i is 3. (you see, the stoichiometry of ions)
Cr₂(SO₄)₃ → 0.10 m
Chromium sulfate is dissociated:
Cr₂(SO₄)₃ → 2Cr³⁺ + 3SO₄⁻²
i = 5
BaI₂ → 0.16 m
BaI₂ → 1Ba²⁺ + 2I⁻
i = 3