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Shalnov [3]
2 years ago
9

A solution of salt and water is 33.0% salt by mass and has a density of 1.50 g/ml. what mass of the salt in grams is in 5.00l of

this solution?
Chemistry
1 answer:
meriva2 years ago
6 0
To answer this item, we solve first for the mass of the solution by multiplying the density by the volume. That is,

      m = (density)(volume)

Substituting the known values,
    m = (1.50 g/mL)(5L)(1000 mL/1L)
      m = 7500 grams

To determine the mass of the salt in the solution, multiply the calculated mass of the solution by the decimal equivalent of the percent salt in the solution.

      m of salt = (7500 g)(0.33)
      m of salt = 2475 grams

<em>Answer: 2475 grams</em>
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A chemical reaction that ______ energy can be used to heat up other substances.
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A galvanic (voltaic) cell consists of an electrode composed of zinc in a 1.0 M zinc ion solution and another electrode composed
MariettaO [177]

Answer:

The E°cell for the galvanic cell is 1.56 V.

Explanation:

A galvanic cell is a device that uses redox reactions to convert chemical energy into electrical energy. The chemical reaction used is always spontaneous.

Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.

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The species that supplies electrons is the reducing agent (that is, it is that species that oxidizes, yielding electrons and increasing its positive charge, or decreasing the negative one causing the reduction of the other species) and the one that gains them is the oxidizing agent ( that is, it is that species that is reduced, capturing electrons and increasing its negative charge, or decreasing its positive charge, causing oxidation of the other species).

The galvanic cell works as follows: In the anodic half-cell oxidations occur, while in the cathodic half-cell reductions occur. The anode electrode, conducts the electrons that are released in the oxidation reaction, to the metallic conductors. These electrical conductors conduct the electrons and carry them to the cathode electrode; the electrons thus enter the cathode half-cell and the reduction takes place in it.

To determine the oxidizing and reducing agent you must first know the reduction potentials. For this you consult the list of standard reduction potentials. In this list you can see that the semi-reactions that occur with their corresponding potentials are:

Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Zn²⁺ + 2 e⁻ ⇒ Zn E° -0.76 V

The species that has the greatest potential for reduction will be the species that will be reduced, that is, it will be the oxidizing agent. In this case, it will be the experience corresponding to silver (Ag). Therefore, to obtain the redox reaction, the half-reaction corresponding to zinc (Zn) must be reversed to be an oxidation, keeping its E ° value constant. Then:

Reduction: Ag⁺ + e⁻ ⇒ Ag E°= 0.80 V

Oxidation: Zn ⇒ Zn²⁺ + 2 e⁻ E° -0.76 V

So: <em>E°cell=Ereduction - Eoxidation</em>

Or what is the same<em> E°cell=Ecathode - Eanode </em>because the reduction always occurs in the cathode and oxidation in the anode.

E°cell=0.80 V - (-0.76) V

<em>E°cell= 1.56 V</em>

Then <u><em>the E°cell for the galvanic cell is 1.56 V.</em></u>

6 0
3 years ago
Solve the fo<br> 1. 750 mL =<br> L
Leokris [45]
759 mililitera is equal to 0.75 liter
7 0
3 years ago
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