Here is the correct option: According to Newton's third law of motion, the balloon is pushed forward as the air is forced out. Newton's third law of motion states that for every action there is an equal and opposite reaction. This means that in every interaction, there is a pair of force that is acting on the interacting objects, the size of the force on the first object is equal to the size of the force on the second object.
Exothermic change. Because the firework when it exploded, released energy in the form of light. In exothermic changes energy is released, and in endothermic changes energy is absorbed.
- This wouldn't be a physical change, but instead a chemical change. A clue that it is a chemical change is that energy was given off.
Answer:
The equilibrium value of [CO] is 1.04 M
Explanation:
Chemical equilibrium is the state to which a spontaneously evolving chemical system, in which a reversible chemical reaction takes place. When this situation is reached, it is observed that the concentrations of substances, both reagents and reaction products, they remain constant over time. That is, the rate of reaction of reagents to products is the same as that of products to reagents.
Reagent concentrations and products in equilibrium are related by the equilibrium constant Kc. Being:
aA + bB ⇔ cC + dD
![Kc=\frac{[C]^{c} *[D]^{d} }{[A]^{a} *[B]^{b} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BC%5D%5E%7Bc%7D%20%2A%5BD%5D%5E%7Bd%7D%20%7D%7B%5BA%5D%5E%7Ba%7D%20%2A%5BB%5D%5E%7Bb%7D%20%7D)
Then this constant Kces equals the multiplication of the concentrations of the products raised to their stoichiometric coefficients between the multiplication of the concentrations of the reactants also raised to their stoichiometric coefficients.
In this case:
![Kc=\frac{[CH_{3}OH ]}{[CO]*[H_{2} ]^{2} }](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BCH_%7B3%7DOH%20%5D%7D%7B%5BCO%5D%2A%5BH_%7B2%7D%20%5D%5E%7B2%7D%20%7D)
You know:
- Kc= 14.5
- [H₂]= 0.322 M
- [CH₃OH] =1.56 M
Replacing:
![14.5=\frac{1.56}{[CO]*0.322^{2} }](https://tex.z-dn.net/?f=14.5%3D%5Cfrac%7B1.56%7D%7B%5BCO%5D%2A0.322%5E%7B2%7D%20%7D)
Solving:
![[CO]=\frac{1.56}{14.5*0.322^{2} }](https://tex.z-dn.net/?f=%5BCO%5D%3D%5Cfrac%7B1.56%7D%7B14.5%2A0.322%5E%7B2%7D%20%7D)
[CO]= 1.04 M
The equilibrium value of [CO] is 1.04 M
The answer would be A) cells!!
<span>Step 1 is to determine the mass of each part
Mass of Ca is 40.08 g
Mass of C is 12.01 g
Mass of O is 16.00 x 3 = 48.00 g
Step 2 is to determine the total mass of the compound
Total mass of CaCO3 is 40.08 + 12.01 + 48.00 = 100.09 g
Step 3 is to determine the % of each part using the following formula:
Mass of part / total mass x 100 =
40.08 / 100.09 x 100 = 40.04 % Ca
12.01 / 100.09 x 100 = 12.00 % C
48.00 / 100.09 x 100 = 47.96 % O
Step 4 is to double check by adding all percentages. If they equal 100, then I probably did it right. :)
40.04
+12.00
+47.96
=100.00</span><span>
</span>
