Batteries work by letting charged ions flow through an electrolyte solution.

The answer is A. Product molecules.

Answer:

H+(aq) + OH-(aq) ⟶ H2O(l)

Explanation:

Step 1: The balanced equation can be presented as

HCN(aq) + KOH(aq) ⟶ H2O(l) + KCN(aq)

H+(aq) + CN-(aq) + K+(aq) + OH-(aq) ⟶ H2O(l) + K+(aq) + CN-(aq)

Step 2: The net ionic equation is formed by removing the spectator ions — those that appear on both sides of the equation — leading us to the final representation:

H+(aq) + OH-(aq) ⟶ H2O(l)

The solution to your inquiry yields P = 17.73 atm. Explanation: The volume V is 250 ml, equivalent to 0.25 liters (L), with a mass of 3.4 g and a temperature of 45°C, which converts to 318°K. We utilize the ideal gas law PV = nRT for the calculations.

2C3H6 (g) + 2NH3 (g) + 3O2 (G) -> 2C3H3N (g) + 6H2O (g)

To begin, this isn't really a chemistry forum, but anyway.

This represents a limiting reagent scenario.

Set it up as a Dimensional Analysis issue.

Begin with your desired outcome.

Your goal is to find the mass of acrylonitrile (C3H3N)

so you should initiate with that (I'll abbreviate Acrylonitrile as ACL for convenience)

(g) = (53 g of ACL/1mol ACL) (2 mols ACL/2 mol C3H6)/ (1mol C3H6/42 grams) (15.0 grams)

If you calculate that, you will find that 15 grams of C3H6 yields 18.9 grams of acrylonitrile produced.

Utilize the same approach for the remaining two reactants.

So, I figured it out, and for

oxygen, I calculated 11.04 grams

and for ammonia, I calculated 15.29 grams

This means that the maximum possible production is 11.04 grams, since to create any additional amount, more O2 would be necessary, but with only 10 grams available, that's the upper limit for this reaction.

The other two reactants are in excess.

Please rate as brainliest!

684 kcal. One mole of glucose weighs roughly 180g. Given that 1g of glucose releases 3.8 kcal, we calculate for 1 mole of glucose: 180g -> 180g * 3.8 kcal/g = 684 kcal.