What can the bromine be classified as

Answers

Answer 1

Answer: I think maybe a liquid if thats what you mean but it can easily become a gas because it evaporates eaisily .

Related Questions

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If there are 4.5 x 1020 atoms of Silver are present, then how many moles of Silver are there? Question 1 Answer a. 0.081 moles b. 12.4 moles c. 0.0007 moles d. 1,338 moles
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The first person to use the term "atom" was _____. Dalton Democritus Aristotle the Egyptians

Answers

the term atom comes from Greek. Democritus was the first person to use the name.

Democritus. Is. The. Correct. Answer I. Hope this will be helpful

In performing chest compressions on a child, the breastbone is compressed to the depth of how many inches? a) 1/2 b) 1 1/2 c) 2. d) 2½

Answers

Answer: A is correct

Explanation: sorry if its wrong

Based on the sign of E cell, classify these reactions as spontaneous or non spontaneous as written.? assume standard conditions. Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) S (s) (nonspontaneous)? Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq) (nonspontaneous)? 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq) (spontaneous?) Are these correct?

Answers

A electrochemical reaction is said to be spontaneous, if $E^(0) cell is positive.$

Answer 1:
Consider reaction: Ni^2+ (aq) + S^2- (aq) ----> + Ni (s) + S (s)

The cell representation of above reaction is given by;
   $S^(2-)/S // Ni^(2+)/Ni$

Hence, $E^(0)cell = E^(0) Ni^(2+/Ni) - E^(0) S/S^(2-)$
we know that, ${E^(0) Ni^(2+)/Ni = -0.25 v$
and ${E^(0) S/ S^(2-) = -0.47 v$

Therefore, $E^(0) cell$ = - 0.25 - (-0.47) = 0.22 v

Since, $E^(0) cell$ is positive, hence cell reaction is spontaneous
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Answer 2:
Consider reaction: Pb^2+ (aq) +H2 (g) ----> Pb (s) +2H^+ (aq)

The cell representation of above reaction is given by;
   $H_(2) / H^(+) // Pb^(2+) /Pb$

Hence, $E^(0)cell = E^(0) Pb/Pb^(2+) - E^(0) H_(2)/H^(+)$
we know that, ${E^(0) Pb^(2+)/Pb = -0.126 v$
and ${E^(0) H_(2)/ H^(+) = -0 v$

Therefore, $E^(0) cell$ = - 0.126 - 0 = -0.126 v

Since,   $E^(0) cell$ is negative, hence cell reaction is non-spontaneous.

....................................................................................................................

Answer 3:
Consider reaction: 2Ag^+ (aq) + Cr(s) ---> 2 Ag (s) +Cr^2+ (aq)

The cell representation of above reaction is given by;
   $Cr/Cr^(2+) // Ag^(+)/Ag$

Hence, $E^(0)cell = E^(0) Ag^(+)/Ag - E^(0) Cr/Cr^(2+)$
we know that, ${E^(0) Ag^(+)/Ag = -0.22 v$
and ${E^(0) Cr/ Cr^(2+) = -0.913 v$

Therefore, $E^(0) cell$ = - 0.22 - (-0.913) = 0.693 v

Since,   $E^(0) cell$ is positive, hence cell reaction is spontaneous

Answer: $Ni^(2+)(aq)+S^(2-)(aq)\rightarrow Ni(s)+S(s)$ : non spontaneous

$Pb^(2+)(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)$ : non spontaneous

$2Ag^(+)(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^(2+)(aq)$ : spontaneous

Explanation:

a) $Ni^(2+)(aq)+S^(2-)(aq)\rightarrow Ni(s)+S(s)$

Here S undergoes oxidation by loss of electrons, thus act as anode. Ni undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_(cathode)- E^0_(anode)$

Where both $E^0$ are standard reduction potentials.

$E^0_([Ni^(2+)/Ni])=-0.25V$

$E^0_([S^(2-)/S])=0.407VV$

$E^0=E^0_([Ni^(2+)/Ni])- E^0_([S^(2-)/S])$

$E^0=-0.25-(0.407V)-0.657V$

As value of $E^0$ is negative, the reaction is non spontaneous.

b) $Pb^(2+)(aq)+H_2(g)\rightarrow Pb(s)+2H^+(aq)$

Here Hydrogen undergoes oxidation by loss of electrons, thus act as anode. Pb undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_(cathode)- E^0_(anode)$

Where both $E^0$ are standard reduction potentials.

$E^0_([Pb^(2+)/Pb])=-0.13$

$E^0_([H^(+)/H_2])=0V$

$E^0=E^0_([Pb^(2+)/Pb])- E^0_([H^(+)/H_2])$

$E^0=-0.13-(0V)=-0.13V$

As value of $E^0$ is negative, the reaction is non spontaneous.

c) $2Ag^(+)(aq)+Cr(s)\rightarrow 2Ag(s)+Cr^(2+)(aq)$

Here Cr undergoes oxidation by loss of electrons, thus act as anode. Ag undergoes reduction by gain of electrons and thus act as cathode.

$E^0=E^0_(cathode)- E^0_(anode)$

Where both $E^0$ are standard reduction potentials.

$E^0_([Ag^(+)/Ag])=+0.80V$

$E^0_([Cr^(2+)/Cr])=-0.913V$

$E^0=E^0_([Ag^(+)/Ag])- E^0_([Cr^(2+)/Cr])$

$E^0=+0.80-(-0.913V)=1.713V$

As value of $E^0$ is positive, the reaction is spontaneous.

In a reaction, chlorine accepts an electron to form Cl-. What is true about the atomic size of the chlorine ion?

Answers

Answer is: The atomic size of the chlorine ion is larger than the size of the chlorine atom.

Covalent radii of chlorine atom (Cl) is 0.099 nm and ionic radii of chlorine anion (Cl⁻) is 0.181 nm.

Difference between an chlorine atom and chlorine anion is the number of electrons that surround the nucleus.

Chlorine atom has 17 electrons and chlorine anion has 18 electrons.

When a neutral atom that has a charge of negative absorbs an electron, this tends to lessen the size of the ion. The larger the negative charge of the negatively charged element, the smaller the size of the ion will be. Hence the answer is that the ion will have a smaller size than the neutral atom

A 25.5 liter balloon holding 3.5 moles of carbon dioxide leaks. If we are able to determine that 1.9 moles of carbon dioxide escaped before the container could be sealed, what is the new volume of the container?

Answers

Answer: The new volume of the container is 13.8 L

Explanation:

To calculate the new volume of the system, we use the equation given by Avogadro's Law. This law states that volume of the gas is directly proportional to the number of moles of the gas at constant pressure.

Mathematically,

$(V_1)/(n_1)=(V_2)/(n_2)$

where,

$V_1\text{ and }n_1$ are the initial volume and moles of the gas.

$V_2\text{ and }n_2$ are the final volume and moles of the gas.

We are given:

$V_1=25.5L\nn_1=3.5\nV_2=?\nn_2=1.9$

Putting values in above equation, we get:

$(25.5)/(3.5)=(V_2)/(1.9)\n\nV_2=13.8L$

Thus the new volume of the container is 13.8 L

Which scientist developed the equation of wave mechanics?

Answers

The scientist that developed the equation of wave mechanics is Erwin Schrodinger that combined the works of Bohr and de Broglie. The electron was a three dimensional wave circling the nucleus in a whole number of wavelengths allowing the waveform to repeat itself as a stable standing wave (it can absorb energy form a nearby source which is oscillating at a proper frequency) representing the energy levels of Bohr model. He also said that Broglie’s model was correct about the matter waves and the electrons are located in the atomic space according to standing wave frequencies.

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