Homogeneous
Hope this helps!!!
Answer:
1.053×10²⁴ atoms of gold
Explanation:
Hello,
Gold nugget are usually the natural occurring gold and they contain 85% - 90% weight of pure gold.
In this question, we're required to find the number of atoms in 344.75g of a gold nugget.
We can use mole concept relationship between Avogadro's number and molar mass.
1 mole = molar mass
Molar mass of gold = 197 g/mol
1 mole = Avogadro's number = 6.022 × 10²³ atoms
Number of mole = mass / molar mass
Mass = number of mole × molar mass
Mass = 1 × 197
Mass = 197g
197g is present in 6.022×10²³ atoms
344.75g will contain x atoms
x = (344.75 × 6.022×10²³) / 197
X = 1.053×10²⁴ atoms
Therefore 344.75g of gold nugget will contain 1.053×10²⁴ atoms of gold
Answer:

Explanation:
Hello,
In this case, considering that the by-mass percent of water is:

Given such percent and the mass of the sample, we can find the mass of water in grams in the sample by solving for it as shown below:

Best regards.
Answer:- 0.138 M
Solution:- The buffer pH is calculated using Handerson equation:

acts as a weak acid and
as a base which is pretty conjugate base of the weak acid we have.
The acid hase two protons(hydrogen) where as the base has only one proton. So, we could write the equation as:

Phosphoric acid gives protons in three steps. So, the above equation is the second step as the acid has only two protons and the base has one proton.
So, we will use the second pKa value. The acid concentration is given as 0.10 M and we are asked to calculate the concentration of the base to make a buffer of exactly pH 7.00.
Let's plug in the values in the equation:



Taking antilog:


On cross multiply:
[base] = 1.38(0.10)
[base] = 0.138
So, the concentration of the base that is
required to make the buffer is 0.138M.