I think the answer is D. Periods 6-7, Because t<span>he periods that would contain elements with electrons in s, p, d, and f orbitals are periods 6-7.
(I hope this helps you a lot! :) Have a nice day!)</span>
It depends on the system if there are more moles of gas on the left and you increase the pressure the yield of a product will increase as the equilibrium moves to the side with the fewest gas molecules
if the right hand side has more moles of gas and you increase the pressure the yield will decrease as the equilibrium shifts to the left hand side where there are more gas moles
hope that helps
Answer:
8.61 mL of the HCl solution
Explanation:
The reaction that takes place is:
- 2HCl + Mg(OH)₂ → MgCl₂ + 2H₂O
From the given mass of Mg(OH)₂, we can calculate <u>the moles of HCl that are neutralized</u>:
- 4x10² mg = 400 mg = 0.400g
- 0.400g Mg(OH)₂ ÷ 58.32g/1mol = 6.859*10⁻³ mol Mg(OH)₂
- 6.859*10⁻³ mol Mg(OH)₂ *
3.429x10⁻³ mol HCl
Finally, to calculate the volume of an HCl solution, we need both the moles and the concentration. We can <u>calculate the concentration using the pH value</u>:
= [H⁺]
- 0.0398 M = [H⁺] = [HCl] *Because HCl is a strong acid*
Thus, the volume is:
- 0.0398 M = 3.429x10⁻³mol HCl / Volume
- Volume = 8.616x10⁻³ L = 8.62 mL
Answer:
Go on the site named " the chemistry of ice cream ''
Explanation:
https://davidson.weizmann.ac.il/en/online/sciencepanorama/chemistry-ice-cream
Reduction reactions are those reactions that reduce the oxidation number of a substance. Hence, the product side of the reaction must contain excess electrons. The opposite is true for oxidation reactions. When you want to determine the potential difference expressed in volts between the cathode and anode, the equation would be: E,reduction - E,oxidation.
To cancel out the electrons, the e- in the reactions must be in opposite sides. To do this, you reverse the equation with the negative E0, then replacing it with the opposite sign.
Pb(s) --> Pb2+ +2e- E0 = +0.13 V
Ag+ + e- ---> Ag E0 = +0.80 V
Adding up the E0's would yield an overall electric cell potential of +0.93 V.
