Answer:
Explanation:
Bases are the species which furnish hydroxide ions in the solution or is capable of forming bonds with electron deficient species as they are electron rich species. When a base accepts a proton, it changes into a acid which is known as its conjugate acid.
Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations. Only the species which are present in aqueous state dissociate. So, the net ionic equation of aqueous solution of ammonia is shown below as:-
The answer should be D. A rate law needs to be rate equaling the rate constant which is represented as k (make sure you use a lower case k since an upper case K is for equilibrium) times the concentrations of each reactant raised to the power of what ever order it has. (if A was a zero order it would be [A]⁰ and if A was third order it would be [A]³).
Do not get the order the reactants are confused with the coefficients in the chemical equation. (just because the reaction has 2B does not mean the rate law will have [B]². As shown in this example since it is first order therefore being [B] in the rate law)
I hope this helps. Let me know if anything is unclear in the comments.
Answer:
B is the answer
Explanation:
B is the most un reasonable answer to do at the BEGINNING of the process. All of the other answers are things you do at the beginning or, before you start it. "Decide whether they have designed the best possible product" means you FINISHED it already, and we want what the engineers needs to do at the beginning. Common sense actually. Your welcome.
<em>A friendly dude that likes to help</em>
If we have 6.68% NaClO, it is the same as saying--> 6.68 grams NaClO= 100 mL of solution. we can use this as a conversion.
800. mL (6.68 mL/ 100 mL)= 53.4 mL
solution = solute + solvent
solute= NaClO
solvent= H2O
solvent= 800-53.4= 747 mL of H2O
so, we you need 53.4 mL of NaClO and 747 mL of water or 53.4 grams of NaClO and 747 mL of water
Answer: The empirical formula is 
.
Explanation:
Mass of C = 1.71 g
Mass of H = 0.287 g
Step 1 : convert given masses into moles.
Moles of C = 
Moles of H = 
Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.
For C =
For H =
The ratio of C: H = 1: 2
Hence the empirical formula is .
