Answer:
if I'm not mistaken, it is a mixture
To determine the mass of the sample in milligrams in this problem, we use the avogadro's number to convert from atoms to moles, relate the moles of element in the sample to the mole present and the molar mass of the sample. We do as follows:
1.552 x 10^22 atoms H ( 1 mol H / 6.022x10^23 atoms H ) ( 1 mol C2H4Cl2 / 4 mol H ) ( 98.96 g C2H4Cl2 / 1 mol C2H4Cl2 ) = 0.625 g C2H4Cl2 = 625 mg <span>C2H4Cl2</span>
Because the gravity levels decrese and so does the oxygen levels
Answer:
3.33 M
Explanation:
It seems your question is incomplete, however, that same fragment has been found somewhere else in the web:
" <em>A chemist prepares a solution of silver nitrate (AgNO3) by measuring out 85.g of silver nitrate into a 150.mL volumetric flask and filling the flask to the mark with water.</em>
<em>Calculate the concentration in mol/L of the chemist's silver nitrate solution. Be sure your answer has the correct number of significant digits.</em> "
In this case, first we <u>calculate the moles of AgNO₃</u>, using its molecular weight:
- 85.0 g AgNO₃ ÷ 169.87 g/mol = 0.500 mol AgNO₃
Then we<u> convert the 150 mL of the volumetric flask into L</u>:
Finally we <u>divide the moles by the volume</u>:
- 0.500 mol AgNO₃ / 0.150 L = 3.33 M
