Answer:
82.4 s
Explanation:
Find the NUMBEr of half lives...then multiply by 54.3
2.27 = 6.5 (1/2)^n
log (2.27/6.5) / log (1/2) = n = 1.52 half lives
1.52 * 54.3 = 82.4 s
Answer:
Explanation:
3.
Knowns: 100mL of solution; concentration of 0.7M
Unknown: number of moles
Equation: number of moles = volume * concentration
Plug and Chug: number of moles = 100/1000 * 0.7 = 0.07 mole
Final Answer: 0.07mole
2.
Knowns: 5.50L of solution; concentration of 0.400M
Unknown: number of moles
Equation: number of moles = volume * concentration
Plug and Chug: number of moles = 5.5 * 0.4 = 2.20 mole
Final Answer: 2.20 mole
Answer: A 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
Explanation:
Given: 
= 0.20 M, 
= 15.0 mL
= 0.10 M, 
= ?
Formula used is as follows.
Substitute the values into above formula s follows.
![M_{1}V_{1} = M_{2}V_{2}\\0.20 M ]times 15.0 mL = 0.10 M ]times V_{2}\\V_{2} = 30 mL](https://tex.z-dn.net/?f=M_%7B1%7DV_%7B1%7D%20%3D%20M_%7B2%7DV_%7B2%7D%5C%5C0.20%20M%20%5Dtimes%2015.0%20mL%20%3D%200.10%20M%20%5Dtimes%20V_%7B2%7D%5C%5CV_%7B2%7D%20%3D%2030%20mL)
Thus, we can conclude that a 0.20 M solution of HCl with a volume of 15.0 mL is exactly neutralized by the 0.10 M solution of NaOH with 3 mL volume.
A. Combustion since it can’t be nuclear….?
Answer:
The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.
Explanation:
To determine the mass of ascorbic acid knowing the number of moles we use the following formula:
number of moles = mass / molecular weight
mass = number of moles × molecular weight
mass of ascorbic acid = 2.82 × 10⁻⁴ × 176
mass of ascorbic acid = 496 × 10⁻⁴ g = 0.0496 g = 49.6 mg
daily requirement of ascorbic acid = 70 - 90 mg
The quantity of ascorbic acid found in sweet lime of 49.6 mg does not meet the daily requirement.
