Answer:
2K + 2H2O → H2 + 2KOH
Explanation:
Find how many atoms you have on both sides then add 2 to both sides.
Reactant: Products:
K: 1+1=2 K: 1+1=2
H: 2+2=4 H: 3+1=4
O: 1+1=2 O: 1+1=2
Therefore it is balanced. Hope this helps
Answer:
3 × 10⁴ kJ
Explanation:
Step 1: Write the balanced thermochemical equation
C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ
Step 2: Calculate the moles corresponding to 865.9 g of H₂O
The molar mass of H₂O is 18.02 g/mol.
865.9 g × 1 mol/18.02 g = 48.05 mol
Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced
According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.
48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ
Answer:
Waters' boiling point decreases with increases in elevation because of the atmospheric pressure.
For example, the higher in elevation you are. The lower the atmospheric pressure is. In other words heated water reached boiling point quicker.
Answer : The balanced equations will be:

Explanation :
The general rate of reaction is,

Rate of reaction : It is defined as the change in the concentration of any one of the reactants or products per unit time.
The expression for rate of reaction will be :
![\text{Rate of disappearance of A}=-\frac{1}{a}\frac{d[A]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20A%7D%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D)
![\text{Rate of disappearance of B}=-\frac{1}{b}\frac{d[B]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20disappearance%20of%20B%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D)
![\text{Rate of formation of C}=+\frac{1}{c}\frac{d[C]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20C%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D)
![\text{Rate of formation of D}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20formation%20of%20D%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
![Rate=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{b}\frac{d[B]}{dt}=+\frac{1}{c}\frac{d[C]}{dt}=+\frac{1}{d}\frac{d[D]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1%7D%7Ba%7D%5Cfrac%7Bd%5BA%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7Bb%7D%5Cfrac%7Bd%5BB%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bc%7D%5Cfrac%7Bd%5BC%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7Bd%7D%5Cfrac%7Bd%5BD%5D%7D%7Bdt%7D)
From this we conclude that,
In the rate of reaction, A and B are the reactants and C and D are the products.
a, b, c and d are the stoichiometric coefficient of A, B, C and D respectively.
The negative sign along with the reactant terms is used simply to show that the concentration of the reactant is decreasing and positive sign along with the product terms is used simply to show that the concentration of the product is increasing.
Now we have to determine the balanced equations corresponding to the following rate expressions.
![Rate=-\frac{d[CH_4]}{dt}=-\frac{1}{2}\frac{d[O_2]}{dt}=+\frac{1}{2}\frac{d[H_2O]}{dt}=+\frac{d[CO_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7Bd%5BCH_4%5D%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BO_2%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7Bd%5BH_2O%5D%7D%7Bdt%7D%3D%2B%5Cfrac%7Bd%5BCO_2%5D%7D%7Bdt%7D)
The balanced equations will be:
