Each of the following are descriptions of physical properties except C. Flammability
Answer:
See explanation
Explanation:
The molecular equation shows all the compounds involved in the reaction.
The molecular equation is as follows;
2NaF(aq) + Pb(NO3)2(aq) -------> PbF2(s) + 2NaNO3(aq)
The complete ionic equation shows all the ions involved in the reaction
The complete ionic equation;
2Na^+(aq) + 2F^-(aq) + Pb^2+(aq) + 2NO3^-(aq) -------->PbF(s) + 2Na^+(aq) +2NO3^-(aq)
The net Ionic equation shows the ions that actually participated in the reaction
The net ionic equation is;
2F^-(aq) + Pb^2+(aq)--------> PbF(s)
Answer:
The boiling point of HF is <u><em>higher than</em></u> the boiling point of H2, and it is <u><em>higher than</em></u> the boiling point of F2.
Explanation:
In HF, inter- molecule forces will be present between the hydrogen and fluorine atoms. There will be hydrogen bonding present among the hydrogen and fluorine atoms. Hydrogen bonds are strong bonds and hence the boiling point for HF would be high as much energy will be required to break these bonds.
H2 and F2 will only have intra-molecular attractions and there will be no hydrogen bonds present in them. As a result, their boiling point will be lower.
Explanation:
5.00 mol H2O × (1 mol C6H12O6/6 mol H2O)
= 0.833 mol C6H12O6
5.00 mol H2O × (6 mol O2/6 mol H2O)
= 5.00 mol O2
<h3>
Answer:</h3>
8 alpha particles
4 beta particles
<h3>
Explanation:</h3>
<u>We are given;</u>
- Neptunium-237
- Thallium-205
- Neptunium-237 undergoes beta and alpha decay to form Thallium-205.
We are required to determine the number of beta and alpha particles produced to complete the decay series.
- We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
- When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.
In this case;
Neptunium-237 has an atomic number 93, while,
Thallium-205 has an atomic number 81.
Therefore;
²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti
We can get x and y
237 = 4x + y(0) + 205
237-205 = 4x
4x = 32
x = 8
On the other hand;
93 = 2x + (-y) + 81
but x = 8
93 = 16 -y + 81
y = 4
Therefore, the complete decay equation is;
²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti
Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.
