Answer is: because carbon(IV) oxide is released.
For example if baking soda is used for <span>the pizza dough, than on heat carbon(IV) oxide is product of chemical reaction:
2NaHCO</span>₃ → Na₂CO₃ + CO₂ + H₂O.
If we use yeast (<span>single-celled microorganism)</span>, also carbon(IV) oxide produced, because yeast turns sugars into carbon(IV) oxide.
Answer:
2-
Explanation:
For an element to be stable, it must follow the octet rule: an atom will gain, lose or share electrons until its valence shell is complete with 8 electrons.
An element with the valence electron configuration ns²np⁴ has 6 (2+4) electrons in its valence shell. Thus, in order to fulfill the octet, it will gain 2 electrons. As a consequence, it will form an anion with charge 2-.
Answer:
0.51M
Explanation:
Given parameters:
Initial volume of NaBr = 340mL
Initial molarity = 1.5M
Final volume = 1000mL
Unknown:
Final molarity = ?
Solution;
This is a dilution problem whereas the concentration of a compound changes from one to another.
In this kind of problem, we must establish that the number of moles still remains the same.
number of moles initially before diluting = number of moles after dilution
Number of moles = Molarity x volume
Let us find the number of moles;
Number of moles = initial volume x initial molarity
Convert mL to dm³;
1000mL = 1dm³
340mL gives 
= 0.34dm³
Number of moles = initial volume x initial molarity = 0.34 x 1.5 = 0.51moles
Now to find the new molarity/concentration;
Final molarity = 
= 
= 0.51M
We can see a massive drop in molarity this is due to dilution of the initial concentration.
Answer:
1.58×10E18
Explanation:
Since we have the reduction potentials we could make decisions regarding which one will be the anode or cathode. Evidently, bromine having the more positive reduction potential will be the cathode while the iodine will be the anode.
E°cell= 1.07- 0.53= 0.54 V
E°cell= 0.0592/n logK
0.54 = 0.0592/2 logK
logK= 0.54/0.0296
logK= 18.2
K= Antilog (18.2)
K= 1.58×10^18
Answer:
the answer is c
Explanation:
My reasoning is that its most likely
