Explanation:
Given that,
The mass of the object, m = 5 kg
We need to find the maximum potential energy. When the height of the object is maximum it will have the maximum potential energy.
The attached figure shows that the maximum height is 25 m. The formula for the potential energy is given by :
So, when the height is 25 m, the object will have the highest potential energy.
Answer:
No
Explanation:
Filling a beaker 200ml of water is not bad because Each beaker's capacity may vary according to the size of different beaker.
Answer:
![K=K_1*K_2\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Explanation:
Hello there!
In this case, for the given chemical reaction, it turns out firstly necessary to write the equilibrium expression for both reactions 1 and 2:
![K_1=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \\\\K_2=\frac{[CO_2][H_2]}{[CO][H_2O]}](https://tex.z-dn.net/?f=K_1%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%5C%5C%5C%5CK_2%3D%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D)
Now, when we combine them to get the overall expression, we infer these two are multiplied to get:
![K=K_1*K_2\\\\K=\frac{[CO][H_2]^3}{[CH_4][H_2O]} *\frac{[CO_2][H_2]}{[CO][H_2O]}\\\\K=\frac{[H_2]^3[CO_2][H_2]}{[CH_4][H_2O][H_2O]}](https://tex.z-dn.net/?f=K%3DK_1%2AK_2%5C%5C%5C%5CK%3D%5Cfrac%7B%5BCO%5D%5BH_2%5D%5E3%7D%7B%5BCH_4%5D%5BH_2O%5D%7D%20%2A%5Cfrac%7B%5BCO_2%5D%5BH_2%5D%7D%7B%5BCO%5D%5BH_2O%5D%7D%5C%5C%5C%5CK%3D%5Cfrac%7B%5BH_2%5D%5E3%5BCO_2%5D%5BH_2%5D%7D%7B%5BCH_4%5D%5BH_2O%5D%5BH_2O%5D%7D)
Regards!
Answer:
B
Explanation:
cause surface area of solute is not necessary for solution
The correct answer is the second option. <span>Baking soda fizzing in vinegar is an example of a chemical change. This example is a chemical reaction which undergoes a chemical change since new substances are being formed the fizzing of the system represent that gas is being produced by the reaction.</span>
