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3 years ago

Explain what happens to the particles in a substance during a physical change.

2 answers:

4 0

When a physical change occurs, the arrangement of particles within the substance may change, but the atoms in the molecules remain bonded together.

melomori [17]3 years ago

4 0

Answer:

Particles stay the same unless there is a chemical change whether the matter is solid, liquid or gas.

Explanation:

in physical changes no new materials are formed and the particles do not change apart from gaining or losing energy

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What property do the elements of the noble gas family all share?

svet-max [94.6K]

Six elemental gases are composed of such exceptionally stable atoms that they almost never react with other elements. They are the gases that make up Group 0 (the rightmost column) of the periodic table: helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn).

8 0

4 years ago

What two things combine with oxygen to form ground-level ozone?

STALIN [3.7K]

Answer:

c) NOx and VOC

oxides of nitrogen (NOx) and volatile organic compounds (VOC)

Explanation:

4 0

3 years ago

Consider a voltaic cell where the anode half-reaction is Zn(s) → Zn2+(aq) + 2 e− and the cathode half-reaction is Sn2+(aq) + 2 e

notsponge [240]

Answer: The concentration of $Sn^{2+}$ in the cell is $9.0\times 10^{-3}M$

Explanation:

We are given:

Oxidation half reaction: $Zn(s)\rightarrow Zn^{2+}(aq.)+2e^-$ $E^o_{Zn^{2+}/Zn}=-0.76V$

Reduction half reaction: $Sn^{2+}(aq.)+2e^-\rightarrow Sn(s)$ $E^o_{Sn^{2+}/Sn}=-0.136V$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction. Here, fluorine will undergo reduction reaction will get reduced.

Here, tin will undergo reduction reaction and will get reduced.

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=-0.136-(-0.76)=0.624V$

To calculate the EMF of the cell, we use the Nernst equation, which is:

$E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Mn^{2+}]}{[Cu^{2+}]}$

where,

$E_{cell}$ = electrode potential of the cell = 0.660 V

$E^o_{cell}$ = standard electrode potential of the cell = +0.624 V

n = number of electrons exchanged = 2

$[Zn^{2+}]=2.5\times 10^{-3}M$

$[Sn^{2+}]$ = ?

Putting values in above equation, we get:

$0.660=0.624-\frac{0.059}{2}\times \log(\frac{2.5\times 10^{-3}}{[Sn^{2+}})$

$[Sn^{2+}]=9.0\times 10^{-3}M$

Hence, the concentration of $Sn^{2+}$ ions is $9.0\times 10^{-3}M$

3 0

3 years ago

The Law of Multiple Proportions states that ____.

harkovskaia [24]

Answer: when two elements combine, A and B, the ratio of the weight of B combining with A is always a whole number

Explanation:

5 0

3 years ago

Gaseous methane (CH4) will react with gaseous oxygen (O2) to produce gaseous carbon dioxide (CO) and gaseous water (H2O) . Suppo

Leto [7]

Answer:

0 g.

Explanation:

Hello,

In this case, since the reaction between methane and oxygen is:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

If 0.963 g of methane react with 7.5 g of oxygen the first step is to identify the limiting reactant for which we compute the available moles of methane and the moles of methane consumed by the 7.5 g of oxygen:

$n_{CH_4}=0.963gCH_4*\frac{1molCH_4}{16gCH_4}=0.0602molCH_4\\ \\n_{CH_4}^{consumed}=7.5gO_2*\frac{1molO_2}{32gO_2}*\frac{1molCH_4}{2molO_2} =0.117molCH_4$

Thus, since oxygen theoretically consumes more methane than the available, we conclude the methane is the limiting reactant, for which it will be completely consumed, therefore, no remaining methane will be left over.

$left\ over=0g$

Regards.

7 0

3 years ago