Yes they will need to worry cause the chemicals could spread in the water making it explode and kill animals and it would be very bad
Answer:
• One mole of oxygen is equivalent to 16 grams.
→ But at STP, 22.4 dm³ are occupied by 1 mole.

Answer:
c
Explanation:
b and d are out, the variables are changed. a would be a repetition, not a replication. c uses the same method and variables with a different control group
The question is incomplete, complete question is :
Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (
for HF is
.)
[HF] = 0.280 M
Express your answer to two decimal places.
Answer:
The pH of an 0.280 M HF solution is 1.87.
Explanation:3
Initial concentration if HF = c = 0.280 M
Dissociation constant of the HF = 

Initially
c 0 0
At equilibrium :
(c-x) x x
The expression of disassociation constant is given as:
![K_a=\frac{[H^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)


Solving for x, we get:
x = 0.01346 M
So, the concentration of hydrogen ion at equilibrium is :
![[H^+]=x=0.01346 M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3Dx%3D0.01346%20M)
The pH of the solution is ;
![pH=-\log[H^+]=-\log[0.01346 M]=1.87](https://tex.z-dn.net/?f=pH%3D-%5Clog%5BH%5E%2B%5D%3D-%5Clog%5B0.01346%20M%5D%3D1.87)
The pH of an 0.280 M HF solution is 1.87.
Answer:
50 g Sucrose
Explanation:
Step 1: Given data
- Concentration of the solution: 2.5%
Step 2: Calculate the mass of sucrose needed to prepare the solution
The concentration of the solution is 2.5%, that is, there are 2.5 g of sucrose (solute) every 100 g of solution. The mass of sucrose needed to prepare 2000 g of solution is:
2000 g Solution × 2.5 g Sucrose/100 g Solution = 50 g Sucrose