HF and NaF - If the right concentrations of aqueous solutions are present, they can produce a buffer solution.
<h3>What are buffer solutions and how do they differ?</h3>
- The two main categories of buffers are acidic buffer solutions and alkaline buffer solutions.
- Acidic buffers are solutions that contain a weak acid and one of its salts and have a pH below 7.
- For instance, a buffer solution with a pH of roughly 4.75 is made of acetic acid and sodium acetate.
<h3>Describe buffer solution via an example.</h3>
- When a weak acid or a weak base is applied in modest amounts, buffer solutions withstand the pH shift.
- A buffer made of a weak acid and its salt is an example.
- It is a solution of acetic acid and sodium acetate CH3COOH + CH3COONa.
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Answer:
[C₆H₁₂O₆] = 0.139 M
Explanation:
Molarity si defined as a sort of concentration. It indicates the moles of solute that are contained in 1 L of solution.
We can also say, that molarity are the mmoles of solute contained in 1 mL of solution.
For this case, the solute is sugar (glucose). Let's determine M (mmol/mL)
(3.95 g . 1mol / 180g) . (1000 mmol / 1mol) / 158 mL
We determine moles, we convert them to mmoles, we divide by mL
M = 0.139 M
Moles = 3.95 g . 1mol / 180g → 0.0219 mol
We convert mL to L → 158 mL . 1L/1000mL = 0.158L
M = 0.0219 mol / 0.158L = 0.139 M
Since the direction of particle displacement in electromagnetic waves is also perpendicular to the direction of motion, generating the waveform of visible light and other forms of electromagnetic radiation, they are also transverse waves.
In a transverse wave, the displacement is perpendicular to the direction of motion (at an angle of 90 degrees Celsius). The direction of displacement (up and down) in the case of the ocean wave is perpendicular to the direction of wave motion (horizontally along the water), making it a transverse wave.
How far a particle has moved from its original starting position, or, in the case of an ocean wave, how high or low the water is, is measured by its displacement or amplitude.
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<u>Given:</u>
Initial concentration of potassium iodate (KIO3) M1 = 0.31 M
Initial volume of KIO3 (stock solution) V1 = 10 ml
Final volume of KIO3 V2 = 100 ml
<u>To determine:</u>
The final concentration of KIO3 i.e. M2
<u>Explanation:</u>
Use the relation-
M1V1 = M2V2
M2 = M1V1/V2 = 0.31 M * 10 ml/100 ml = 0.031 M
Ans: The concentration of KIO3 after dilution is 0.031 M
D.radiation that’s the right answer
