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olga55 [171]
3 years ago
13

Is PCI3 a good conductor?

Chemistry
1 answer:
AVprozaik [17]3 years ago
5 0

Answer:

Since the ions cannot move, ionic solids are non-conductors of electricity. When the solid is melted to a liquid, however, the ions are free to migrate. Therefore, ionic liquids are good conductors of electric current. A conductor is a substance capable of transmitting electricity and/or heat.

Explanation:

:)

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Balance equation P2O5 + H20 - - - - &gt; H3PO4<br> Help me.....
Archy [21]

Answer:

Search by reactants (P 2O 5, H 2O) and by products (H 3PO 4)

H2O + P2O5 → H3PO4

H2O + HNO3 + P2O5 → H3PO4 + N2O5

8 0
3 years ago
Hsgsfqg hsagdjhagsydt ahsgdytratdw
stiks02 [169]

Answer:

Yes

Explanation:

3 0
2 years ago
B. The following reaction takes place in a basic solution. (7 points)
Natali5045456 [20]

<u>Answer:</u> The final equation has hydroxide ions which indicate that the reaction has occurred in a basic medium.

<u>Explanation:</u>

Redox reaction is defined as the reaction in which oxidation and reduction take place simultaneously.

The oxidation reaction is defined as the reaction in which a chemical species loses electrons in a chemical reaction. It occurs when the oxidation number of a species increases.

A reduction reaction is defined as the reaction in which a chemical species gains electrons in a chemical reaction. It occurs when the oxidation number of a species decreases.

The given redox reaction follows:

MnO_4^-(aq)+NO_2^-(aq)\rightarrow MnO_2(s)+NO_3^-(aq)

To balance the given redox reaction in basic medium, there are few steps to be followed:

  • Writing the given oxidation and reduction half-reactions for the given equation with the correct number of electrons

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-

  • Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions

Oxidation half-reaction: NO_2^-+2OH^-\rightarrow NO_3^-+H_2O+2e^-         ( × 3)

Reduction half-reaction: MnO_4^-+2H_2O+3e^-\rightarrow MnO_2+4OH^-             ( × 2)

The half-reactions now become:

Oxidation half-reaction: 3NO_2^-+6OH^-\rightarrow 3NO_3^-+3H_2O+6e^-

Reduction half-reaction: 2MnO_4^-+4H_2O+3e^-\rightarrow 2MnO_2+8OH^-

  • Add the equations and simplify to get a balanced equation

Overall redox reaction: 3NO_2^-+2MnO_4^-+H_2O\rightarrow 3NO_3^-+2MnO_2+2OH^-

As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution

3 0
3 years ago
What type of radiation is likely to occur when the ratio of protons to neutrons is below the band of stability? a. alpha decay b
Natali [406]
Beta radiation / decay would likely occur when the ratio of protons to neutrons is below the band of stability.
7 0
3 years ago
Can somebody help me, please!
zavuch27 [327]

Answer:

Rubidium-85=61.2

Rubidium-87=24.36

Atomic Mass=85.56 amu

Explanation:

To find the atomic mass, we must multiply the masses of the isotope by the percent abundance, then add.

<u>Rubidium-85 </u>

This isotope has an abundance of 72%.

Convert 72% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 72/100= 0.72      or        72.0 --> 7.2 ---> 0.72

Multiply the mass of the isotope, which is 85, by the abundance as a decimal.

  • mass * decimal abundance= 85* 0.72= 61.2

Rubidium-85=61.2

<u>Rubidium-87</u>

This isotope has an abundance of 28%.

Convert 28% to a decimal. Divide by 100 or move the decimal two places to the left.

  • 28/100= 0.28       or        28.0 --> 2.8 ---> 0.28

Multiply the mass of the isotope, which is 87, by the abundance as a decimal.

  • mass * decimal abundance= 87* 0.28= 24.36

Rubidium-87=24.36

<u>Atomic Mass of Rubidium:</u>

Add the two numbers together.

  • Rb-85 (61.2) and Rb-87 (24.36)
  • 61.2+24.36=85.56 amu
4 0
3 years ago
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