Answer:
746 moles of H2O are been produced from 373 moles of Al.
Explanation:
For every 3 moles of aluminum, you get 6 moles of H2O (double). Therefore, every 373 moles of Al, you will get double as well, that is 746 m.
Answer:
There are 3600 pages to 2 significant figures
Explanation:
Working only with the information given, we have that there are 23 books each with about 31 chapters.
This means that if 1 book has 31 chapter, the 23 science books will have
31 X 23 chapters =713 chapters in total.
If each chapter contains 5 pages, for the 713 chapters we will have a total of 713 X 5 pages = 3565 pages.
Rounding off the answer to 2 significant figures , we are looking to have only two non-zero figures in our answer. To do this, we count two numbers from the left. This will be numbers 3 and 5.
Next, we check the net number after 5 to see if it is greater than 5 if it is, we will have to add the number (1 ) to the 5. This shows that we are approximating upwards, to account for the fact that that number has passed the middle point.
After this, we change the remaining two numbers (6 and 5) to zeroes.
This will give us 3600 pages.
Answer:
B. Write the chemical equation using formulas and symbols.
D. Count the atoms in each substance in the reactants and products,
Explanation:
When balancing chemical equation, it implies that the law of conservation of matter be strictly adhered to.
By so doing, the number of atoms on both sides of the reaction must be the same. Number of atoms in the products and reactants must have the same value.
- It is always a good approach to first write the chemical equation of the reaction.
- This will give an overview of the reacting species and the products being formed.
- Then, go on to count the number atoms in the reactants and product sides of the expression.
- Then balance it using any of the method of balancing chemical equations.
Answer:
a.
![Keq=\frac{[HCO_3^-][OH^-]}{[CO_3^{2-}]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BHCO_3%5E-%5D%5BOH%5E-%5D%7D%7B%5BCO_3%5E%7B2-%7D%5D%7D)
b.
![Keq=[O_2]^3](https://tex.z-dn.net/?f=Keq%3D%5BO_2%5D%5E3)
c.
![Keq=\frac{[H_3O^+][F^-]}{[HF]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BF%5E-%5D%7D%7B%5BHF%5D%7D)
d.
![Keq=\frac{[NH_4^+][OH^-]}{[NH_3]}](https://tex.z-dn.net/?f=Keq%3D%5Cfrac%7B%5BNH_4%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BNH_3%5D%7D)
Explanation:
Hello there!
In this case, for the attached reactions, it turns out possible for us to write the equilibrium expressions by knowing any liquid or solid would be not-included in the equilibrium expression as shown below, with the general form products/reactants:
a.
b.
c.
d.
Regards!
