Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer:
22Ω
Explanation:
Given parameters:
Potential difference = 3.3V
Current = 0.15A
Unknown:
Resistance = ?
Solution:
According to ohm's law, potential difference, current and resistance are related by the expression below;
V = I R
where V is the voltage
I is the current
R is the resistance
3.3 = 0.15 x R
R = 
= 22Ω
B; Seawater mixes with freshwater so the water has intermediate salinity
Explanation:
In an estuary, seawater mixes with freshwater so the water has intermediate salinity. Estuaries are usually located in transitional environments.
- Estuary is the wide part of a river where it nears the sea.
- This is called a transitional zone.
- Water from continental rivers usually fresh are brought in close contact with ocean water that is salty.
- The water here is said to be brackish as it is intermediate between salt and seawater.
- Organisms living in such terrain must be be well adapted to changing salinity.
Learn more:
salinity and density brainly.com/question/10491444
#learnwithBrainly
Explanation:
36.5gm (Molar mass) = 1mole
1 gram = 1/36.5 mole
10grams =(1/36.5) x 10 moles
= 0.027 x 10 moles
=0.27moles
