I believe the answer is true. Hope this helps.
5 x 2 = 10
3 x 3 = 9
10/9, or 1 1/9 is your answer
hope this helps
Answer:
Mass of chemical = 1.5 mg
Explanation:
Step 1: First calculate the concentration of the stock solution required to make the final solution.
Using C1V1 = C2V2
C1 = concentration of the stock solution; V1 = volume of stock solution; C2 = concentration of final solution; V2 = volume of final solution
C1 = C2V2/V1
C1 = (6 * 25)/ 0.1
C1 = 1500 ng/μL = 1.5 μg/μL
Step 2: Mass of chemical added:
Mass of sample = concentration * volume
Concentration of stock = 1.5 μg/μL; volume of stock = 10 mL = 10^6 μL
Mass of stock = 1.5 μg/μL * 10^6 μL = 1.5 * 10^6 μg = 1.5 mg
Therefore, mass of sample = 1.5 mg
Answer:
Na₂₆F₁₁
Explanation:
We find the moles of the substance assuming 100 g of the substance is present. Why do we take 100 g? Because then the percent of sodium/fluorine, would be the g of sodium/fluorine respectively:
74.186 g Sodium | 1 mol Sodium/23 g => 3.2255 mol Na
25.814 g Fluorine | 1 mol Fluorine/19 g => 1.3586 mol F
Divide each by smallest number of moles:
3.2255/1.3586 = 2.37
1.3586/1.3586 = 1
Multiply by common number to get a smallest whole number:
2.37*11 = 26,
1*11 = 11
The empirical formula is Na₂₆F₁₁
Answer:
18.91 ml
Explanation:
Initial volume of base=2.04 ml
Final volume of base = 20.95 ml
Volume of base used= 20.95 - 2.04 = 18.91 ml
Note that the volume of base used is obtained as the difference between the final and initial volume of base, hence the answer given above.
