Answer:
The volume of the sample given is 850 ml, the density given is 0.79 gram per cm. Now the weight of the sample will be,
Weight = volume × density = 850 × 0.79
= 671.5 grams
Weight of the suspended solids given is 0.001 gram
The concentration of the sample can be determined by using the formula,
Concentration = wt. of sample/volume
= [671.5 - 0.001) 10³ mg / 0.85 L
= 789998.82 mg/L or 789998.82 ppm
Now the concentration of suspended solids is.
Css = 0.001 × 10³ mg / 0.85 L = 1.1764 mg per L or 1.1764 ppm
Answer:
3.6 × 10²⁴ atoms of O
Explanation:
Let's consider the molecular formula of silver nitrate: AgNO₃.
We can establish the following relations:
- 1 mol of AgNO₃ has 6.02 × 10²³ molecules of AgNO₃ (Avogadro's number).
- 1 molecule of AgNO₃ has 3 atoms of oxygen.
The atoms of oxygen in 2.0 moles of silver nitrate are:
Answer:
i i cant help youu withh thatt srry i i juss dont knoww whatt to say !!!!! GO TO SOCRATICCC ..!!!!!!!!!!!!!$$!$!
