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1 answer:

8 0

Calcium reacts gently with water to give hydrogen and calcium hydroxide, which is only slightly soluble, thus slows down the reaction.

It will be assumed that hydrochloric acid used is a dilute aqueous solution.

However, calcium reacts with hydrochloric acid to give calcium chloride which is readily soluble in water, and hydrogen, being a typical reaction of relatively active metals with acids.
Ca(s) + 2HCl(aq) -> CaCl2(aq) +H2(g) ↑ + heat

The clues that it is a chemical reaction could be:
- formation of a new substance, gaseous hydrogen
- disappearance of a metallic solid in the solution
- heat formed during the vigorous reaction.

As silver is below hydrogen in the electrochemical series, it will not be expected to react with dilute hydrocloric acid. (however, it dissolves in oxidizing acid such as nitric acid, but not displacing hydrogen as a product).

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A solution is prepared by dissolving 0.23 mol of hypochlorous acid and 0.27 mol of sodium hypochlorite in water sufficient to yi

finlep [7]

Answer:

hypochlorite ion

Explanation:

The hypochlorous acid, HClO, is a weak acid with Ka = 1.36x10⁻³, when this acid is in solution with its conjugate base, ClO⁻ (From sodium hypochlorite, NaClO) a buffer is produced. When a strong acid as HCl is added, the reaction that occurs is:

HCl + ClO⁻ → HClO + Cl⁻.

Where more hypochlorous acid is produced.

That means, the HCl reacts with the hypochlorite ion present in solution

3 0

3 years ago

Five million gallons per day (MGD) of wastewater, with a concentration of 10.0 mg/L of a conservative pollutant, is released int

hjlf

Answer:

a) The concentration in ppm (mg/L) is 5.3 downstream the release point.

b) Per day pass 137.6 pounds of pollutant.

Explanation:

The first step is to convert Million Gallons per Day (MGD) to Liters per day (L/d). In that sense, it is possible to calculate with data given previously in the problem.

Million Gallons per day $1 MGD = 3785411.8 litre/day = 3785411.8 L/d$

$F_1 = 5 MGD (\frac{3785411.8 L/d}{1MGD} ) = 18927059 L/d\\F_2 =10 MGD (\frac{3785411.8 L/d}{1MGD} )= 37854118 L/d$

We have one flow of wastewater released into a stream.

First flow is F1 =5 MGD with a concentration of C1 =10.0 mg/L.

Second flow is F2 =10 MGD with a concentration of C2 =3.0 mg/L.

After both of them are mixed, the final concentration will be between 3.0 and 10.0 mg/L. To calculate the final concentration, we can calculate the mass of pollutant in total, adding first and Second flow pollutant, and dividing in total flow. Total flow is the sum of first and second flow. It is shown in the following expression:

$C_f = \frac{F1*C1 +F2*C2}{F1 +F2}$