Answer:
b. 7.5 x 10^-3
Explanation:
To solve this problem we need to keep in mind the <em>definition of molarity</em>:
- Molarity = moles of solute / liters of solution
With the above information in mind it is possible to calculate the moles of solute, given the volume (10 mL) and concentration (0.75 M) of the solution:
- First we<u> convert 10 mL to L</u> ⇒ 10 mL / 1000 = 0.01 L
Then we <u>calculate the moles of AgNO₃</u>:
- moles of solute = Molarity * Liters of solution
- 0.01 L * 0.75 M = 7.5x10⁻³ mol AgNO₃
<em>One mole of AgNO₃ contains one mole of Ag⁺</em>, thus the number of Ag⁺ moles is also 7.5x10⁻³.
The quality of being easily dissolved in liquid.
Hope this help and Happy holidays
Answer:
pH = 5.76
Explanation:
We can solve this problem by using<em> Henderson-Hasselbach's equation</em>:
pH = pKa + log![\frac{[SodiumAcetate]}{[AceticAcid]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BSodiumAcetate%5D%7D%7B%5BAceticAcid%5D%7D)
We are already know all the required information, thus we<u> input the data given by the problem</u>:
pH = 4.76 + log(20/2)
And finally <u>calculate the pH</u>:
pH = 5.76
The pH of that acetic acid solution is 5.76.
An anode is an electrode, it can be a metal or another conductor. in an electrochemical cell that is polarized if an electric current flows into it. Electric current flows opposite to the direction of movement of electrons. In electrochemical processes, both galvanic cells (batteries) and electrolysis cells, anodes undergo oxidation.
In contrast to an Anode, a Cathode is an electrode pole in an electrochemical cell that is polarized if this pole is positively charged (so that an electric current will flow out of it, or the movement of electrons will enter this pole).
In galvanic cells or power plants (batteries), the anode is the negative pole. The electrode will release electrons towards the circuit and hence an electric current flows into this electrode and makes it an anode and negative.
Answer:
A. 15859.2 L or 15900 L
B. 0.629 mol
Explanation:
At STP, one mole is equal to approximately 22.4 L
L or mL is volume, so you are attempting to solve for L or mL.
A.
708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)
B.
(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.
