Answer:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Explanation:
We'll begin by writing the dissociation equation for aqueous AgNO₃ and KI.
Aqueous AgNO₃ and KI will dissociate in solution as follow:
AgNO₃ (aq) —> Ag⁺(aq) + NO₃¯ (aq)
KI (aq) —> K⁺(aq) + I¯(aq)
Aqueous AgNO₃ and KI will react as follow:
AgNO₃ (aq) + KI (aq) —>
Ag⁺(aq) + NO₃¯ (aq) + K⁺ (aq) + I¯(aq) —> AgI (s) + K⁺ (aq) + NO₃¯ (aq)
Cancel out the spectator ions (i.e ions that appears on both sides of the equation) to obtain the net ionic equation. The spectator ions are K⁺ and NO₃¯.
Thus, the net ionic equation is:
Ag⁺ (aq) + I¯ (aq) —> AgI (s)
Answer : The compound contains the highest percentage of lead (by mass) is, PbS.
Explanation :
To calculate the percentage of lead in sample, we use the equation:

<u>For
:</u>
Mass of
= 239.3 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:

The percentage of lead in the
is 86.58 %.
<u>For
:</u>
Mass of
= 267.2 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:

The percentage of lead in the
is 77.55 %.
<u>For
:</u>
Mass of
= 259.7 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:

The percentage of lead in the
is 79.78 %.
<u>For
:</u>
Mass of
= 545.3 g
Mass of Pb = 207.2 g
Putting values in above equation, we get:

The percentage of lead in the
is 37.99 %.
Hence, from this we conclude that, the compound PbS contains the highest percentage of lead (by mass).
Answer:
Group 1 and 2 elements
Explanation:
Nitrogen, a non-metal will form ionic bonds with most group 1 and group 2 metals on the periodic table.
How does ionic bonds form?
- They are bonds formed between a highly electronegative specie and one with very low electronegativity.
- As such, ionic bonds forms between metals and non-metals
- In this bond type, the metal due to its electropositive nature will transfer electrons to the non-metals for it to gain.
- The non-metals becomes negatively charged as the metal is positively charged.
- The electrostatic attraction between the two specie leads to the formation of ionic bonds.
Most metals in group 1 and 2 fits in this description. Some of them are calcium, magnesium, lithium, Barium e.t.c.
It mostly favors group 2 metals.
Answer:
a) Germanium = 5.76 x 〖10〗^11 〖cm〗^(-3) , Semiconductor is n-type.
b) Silicon = 2.25 x 〖10〗^5 〖cm〗^(-3) , Semiconductor is n-type.
For clear view of the answers: Please refer to calculation 5 in the attachments section.
Explanation:
So, in order to find out the concentration of holes and electrons in a sample of germanium and silicon which have the concentration of donor atoms equals to 〖10〗^15 〖cm〗^(-3). We first need to find out the intrinsic carrier concentration of silicon and germanium at room temperature (T= 300K).
Here is the formula to calculate intrinsic carrier concentration: For calculation please refer to calculation 1:
So, till now we have calculated the intrinsic carrier concentration for germanium and silicon. Now, in this question we have been given donor concentration (N_d) (N subscript d), but if donor concentration is much greater than the intrinsic concentration then we can write: Please refer to calculation 2.
So, now we have got the concentration of electrons in both germanium and silicon. Now, we have to find out the concentration of holes in germanium and silicon (p_o). (p subscript o)
Equation to find out hole concentration: Please refer to calculation 3. and Calculation 4. in the attachment section.
Good Luck Everyone! Hope you will understand.