Answer:
41.63g
Explanation:
Given parameters:
Volume of CaCl₂ = 500mL = 0.5L
Concentration = 0.75mol/L
Unknown:
Mass of the solute needed = ?
Solution:
The mass of the solute can be derived using the expression below;
Mass = number of moles x molar mass
But,
Number of moles = Concentration x Volume
So;
Mass = Concentration x Volume x molar mas
Molar mass of CaCl₂ = 40 + 2(35.5) = 111g/mol
Mass = 0.75 x 0.5 x 111 = 41.63g
Answer:
equivalent exchange forces cancel out but the substances are affected around the area
Explanation:
<span>A 1 molar solution is the molecular weight in grams in 1 litre of water, so a 3.5 molar solution would be 58.44g multiplied by 3.5, which is 204.54g in 1L.</span>
Answer:
7.08
Explanation:
To solve this problem we'll use the <em>Henderson-Hasselbach equation</em>:
- pH = pka + log
![\frac{[A^-]}{[HA]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BA%5E-%5D%7D%7B%5BHA%5D%7D)
Where is the ratio of [sodium formate]/[formic acid] and pka is equal to -log(Ka), meaning that:
- pka = -log (1.8x10⁻⁴) = 3.74
We<u> input the data</u>:
- 4.59 = 3.74 + log
And<u> solve for </u>:
- 0.85 = log
=
- = 7.08
