Answer:
Search by reactants (P 2O 5, H 2O) and by products (H 3PO 4)
H2O + P2O5 → H3PO4
H2O + HNO3 + P2O5 → H3PO4 + N2O5
<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:

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: 
Reduction half-reaction: 
- Multiply each half-reaction by the correct number in order to balance charges for the two half-reactions
Oxidation half-reaction:
( × 3)
Reduction half-reaction:
( × 2)
The half-reactions now become:
Oxidation half-reaction: 
Reduction half-reaction: 
- Add the equations and simplify to get a balanced equation
Overall redox reaction: 
As we can see that in the overall redox reaction, hydroxide ions are released in the solution. Thus, making it a basic solution
Beta radiation / decay would likely occur when the ratio of protons to neutrons is below the band of stability.
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)