27 I think beacuse if you subtract
Answer:
17.934 kg of water
Explanation:
If balanced equation is not given; this format can come in handy.
For any alkane of the type : CₙH₂ₙ₊₂ , it's combustion reaction will follow:
2CₙH₂ₙ₊₂ + (3n+1) O₂ → (2n)CO₂ + 2(n+1) H₂O
For butane:
2C₄H₁₀(g) + 13O₂(g) → 8CO₂(g) + 10H₂O(l)
2 moles of butane gives 10 moles of water.
1 mol of any substance has Avogadro number(N) of molecules in it( 6.022 x 10²³)
Mass of 1 mole of any substance is equal to it's molar mass
So, if 2 x N molecules of butane gives 10 x 18 g of water.
Then 1.2 x 10²⁶ molecules will give:
= 17.934 x 10³ g of water
= 17.934 kg of water
The pH scale is used to measure the degree of acidity or alkalinity of a solution. The scale runs from 0 (very acidic solutions can have a negative pH) to 14 (very alkaline solutions can have a pH higher than this), while a neutral liquid such as pure water has a pH of 7. The pH is linked to the concentration of hydrogen ions (H +) in the solution. Diluting an acid or alkali affects the concentration of H +<span> ions in a solution and therefore affects the pH. In this activity, we will investigate how diluting an acid or alkali affects the pH.
Hope this helps:D
Have a great rest of a brainly day!</span>
Yes, free electrons appear in balanced redox reaction equations. However, this is only true for half-reactions. This is because redox reactions primarily involve the transfer of electrons, which are better visualized if explicitly shown in the balanced reactions. In reduction reactions, electrons are placed on the left side of the equation. Oxidation reactions show electrons on the right side of the equation.
Explanation:
A half reaction is either the chemical reaction or reduction reaction part of an oxidoreduction reaction. A half reaction is obtained by considering the amendment in chemical reaction states of individual substances concerned within the oxidoreduction reaction. Half-reactions are usually used as a way of leveling oxidoreduction reactions.The half-reaction on the anode, wherever chemical reaction happens, is Zn(s) = Zn2+ (aq) + (2e-).
The metal loses 2 electrons to create Zn2+. The half-reaction on the cathode wherever reduction happens is Cu2+ (aq) + 2e- = Cu(s).
Here, the copper ions gain electrons and become solid copper.
