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nexus9112 [7]
4 years ago
7

The change of energy from one form to another results in the __ of some heat.

Chemistry
1 answer:
beks73 [17]4 years ago
3 0

The answer is Vaporizing

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In an experiment, students were given an unknown mineral. The unknown mineral was placed in 150 ml of water. Once in the water,
Dahasolnce [82]

Answer:

<h2>15 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

But from the question

volume = final volume of water - initial volume of water

volume = 165 - 150 = 15 mL

We have

density =  \frac{225}{15}  = 15 \\

We have the final answer as

<h3>15 g/mL</h3>

Hope this helps you

5 0
3 years ago
Most favourable conditions for electrovalency are
anygoal [31]

Answer:

I think

Explanation:

low charge of ions, large cation and small anion

7 0
4 years ago
Calculate the concentration of formate in a 100mm solution of formic acid at ph 4.15. The pka for formic acid is 3.75
MissTica

The molarity of formic acid is 100 mM or 100\times 10^{-3}M. The dissociation reaction of formic acid is as follows:

HCOOH\leftrightharpoons HCOO^{-}+H^{+}

The expression for dissociation constant of the reaction will be:

K_{a}=\frac{[HCOO^{-}][H^{+}]}{[HCOOH]}

Rearranging,

[HCOO^{-}]=\frac{K_{a}[HCOOH]}{[H^{+}]}

Here, pH of solution is 4.15 thus, concentration of hydrogen ion will be:

[H^{+}]=10^{-pH}=10^{-4.15}=7.08\times 10^{-5}M

Similarly, pK_{a}=3.75 thus,

[K_{a}=10^{-pK_{a}}=10^{-3.75}=1.78\times 10^{-4}M

Putting the values,

[HCOO^{-}]=\frac{(1.78\times 10^{-4}M)(100\times 10^{-3}M)}{(7.08\times 10^{-5}M}=0.2511 M

Therefore, the concentration of formate will be 0.2511 M.

4 0
3 years ago
PH3, explain why there is a partial positive on hydrogen and partial negative on phosphorus.
Paraphin [41]

Answer:

Phosphorus is more electronegative than hydrogen

Explanation:

Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons towards itself thereby making a molecule to be polar. The Pauling scale is the most commonly used to measure electronegativity. Fluorine (the most electronegative element) is assigned a value of 4.0 on the Pauling's scale, and values range down to caesium and francium which are the least electronegative elements.

Electronegativity increases from left to right across the periodic table (across the period) hence, phosphorus is far more electronegative than hydrogen. Being more electronegative than hydrogen, phosphorus attracts the bonding electron pair of the P-H bond closer to itself than hydrogen. Since the electrons of the bond are closer to phosphorus than hydrogen, the phosphorus atom acquires a partial negative charge while the hydrogen atom acquires a partial positive charge.

3 0
3 years ago
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring
Lady_Fox [76]

Answer:

The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)

Explanation:

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.

An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.

In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .

Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.

The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.

In this case you have the reaction:

3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)

In this case the following half-reactions occur:

Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V

Semi-reaction of reduction (an atom or group of atoms gains electrons, increasing its negative charges): Cl₂(g) + 2 e- --> 2 Cl-(aq); E⁰=1.36 V

In an electrochemical cell at 25°C  the potentials of the  semi-reactions are usually measured  in the sense of reduction  and generally the standard potential between both electrochemical cells will be:

E^{0} =E^{0} _{reduction} -E^{0} _{oxidation}

E⁰=1.36 V - (-0.04 V)

E⁰=1.36 V + 0.04 V

<em>E⁰=1.40 V</em>

<em><u>The standard cell potential is 1.40 V. The correct option is the option  D (+1.40 V)</u></em>

7 0
3 years ago
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