Answer : The value of 
for the reaction is, -565.6 kJ
Explanation :
First we have to calculate the molar mass of CO.
Molar mass CO = Atomic mass of C + Atomic mass of O = 12 + 16 = 28 g/mole
Now we have to calculate the moles of CO.
Now we have to calculate the value of for the reaction.
The balanced equation will be,
From the balanced chemical reaction we conclude that,
As, 
of CO release heat = 10.1 kJ
So, 2 mole of CO release heat = 
Therefore, the value of for the reaction is, -565.6 kJ (The negative sign indicates the amount of energy is released)
D, since it can hold up to 10 electrons
A chemical reaction is the process in which atoms present in the starting substances rearrange to give new chemical combinations present in the substances formed by the reaction. These starting substances of a chemical reaction are called the reactants, and the new substances that result are called the products.
A) all mechanical waves need a medium to travel through.
This is true because mechanical waves move in a wave like manner at the atomic level. This is why an table vibrates when you put a speaker on top of it .
Answer:
Increase in CO2 (g) over time.
No NaHCO3 (s) will be left after a time
Explanation:
The reaction, shown below;
2NaHCO3(s) → Na2CO3(s)+CO2(g)+H2O(ℓ) is a decomposition reaction. A decomposition reaction is a kind of chemical reaction in which a given chemical specie breaks up to give other chemical species. Decomposition may be induced by heat or light.
Usually, there is only one reactant in a decomposition reaction; the specie that disintegrates into the products. This reactant usually decreases in concentration steadily because it is converted into products. This is why the mass of NaHCO3(s) in the system continues to decrease steadily until it finally falls to zero.
Conversely, the concentration (for aqueous) or volume (for gases) or mass (for solid) products of the reaction increases steadily as the reaction progresses. This explains why the volume of CO2 in the system will steadily increase over time.
