Answer: Volume of 1g of pure gold 
Given;
Density of a pure gold=19.3 
Mass of a pure gold =1g
To find:
Volume of 1g of pure gold
Solution:
According to the formula,
Density= Mass/Volume
Where 
=density of pure gold
m=mass of pure gold
v=volume of pure gold
From the above equation volume can be calculated as
Substitute the values of mass and density value in the above equation
Result:
Thus the volume of 1g of pure gold is
The volume of 15.7 M H2SO4 is required to prepare 12.0 L of 0.156 M sulfuric acid is 0.12 L
<h3>Data obtained from the question</h3>
From the question given above, the following data were obtained:
- Molarity of stock solution (M₁) = 15.7 M
- Volume of diluted solution (V₂) = 12 L
- Molarity of diluted solution (M₂) = 0.156 M
- Volume of stock solution needed (V₁) = ?
<h3>How to determine the volume of the stock solution needed</h3>
The volume of the stock solution needed can be obtained by using the dilution formula as shown below:
M₁V₁ = M₂V₂
15.7 × V₁ = 0.156 × 12
15.7 × V₁ = 1.872
Divide both side by 15.7
V₁ = 1.872 / 15.7
V₁ = 0.12 L
Thus, the volume of the stock solution needed to prepare the solution is 0.12 L
Learn more about dilution:
brainly.com/question/15022582
#SPJ1
Answer:
Option B
Explanation:
Avagadro's hypothesis showed that at constant temperature and pressure equal volume of all gases contains equal no of molecules.
Avagadro's constant is known as 6.022×10^23
Work Done = force x displacement. So in this case the 15N is the force (because weight is a force) and 0.60m is the displacement. Therefore 15 x 0.6 = 9 Joules of work done (btw, work done can also be referred to as energy transferred)
