given days = 42 days
So, number of half life, n = 42/14.3 = 2.93
So, after 42 days, the mass of sodium phosphate sample left = original mass
x (1/2)^n,
= 157 mg x (1/2)^2.93 = 20.60 g of Na3PO4 after 42 days
Formula mass of Sodium phosphate = 3(23) + 32 + 4(16) amu = 165 amu
165 g of sodium phosphate contains 32 g of Phosphorus
20.60 g ------------------------------? = 20.60 x 32/ 165 =3.99g of P32
Answer:
<u></u>
<u></u>
- <u>b. See the description below</u>
Explanation:
<u><em>a. Volume of 0.400 M CuSO₄(aq) required for the preparation</em></u>
In dissolutions, since the number of moles of solute is constant, the equation is:

Substitute and solve for V₁


<u><em>b. Briefly describe the essential steps to most accurately prepare the 0.150 M CuSO₄(aq) from the 0.400 M CuSO₄(aq)</em></u>
You will use the stock solution, the funnel, the buret, and distilled water.
i) Using the funnel, fill in the buret with with 50 ml of the stock solution, i.e. the 100. ml of 0.400 M CuSO₄(aq) solution.
ii) Pour 37.5 ml of the stock solution from the burete into the volumetric flask.
iii) Carefully add disitlled water to the 37.5ml of the stock solution in the volumetric flask until the mark (50 ml) on the volumetric flask.
iv) Put the stopper and rotate the volumetric flask to homegenize the solution.
Answer:
hope this will help you......
The branch of chemistry dealing with the physical changes associated with chemical reactions
<em>the</em><em> </em><em>number</em><em> of</em><em> </em><em>nitrogen</em><em> </em><em>atoms</em><em> </em><em>in</em><em> </em><em>the </em><em>compound</em><em> </em><em>is</em><em> </em><em>two</em>