Mass of lead (II) chromate is 51 g. The molecular formula is 
and its molar mass is 323.2 g/mol
Number of moles can be calculated using the following formula:
Here, m is mass and M is molar mass.
Putting the values,
Therefore, number of moles of lead (II) chromate will be 0.1578 mol.
Answer is: selenium (Se).
1) electron configuration: ₃₄Se 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp⁴.
2) ₃₃As 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp³.
3) ₃₆Kr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp⁶.
4) ₃₁Ga 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4sp¹.
Valence electrons of selenium are 4s²4sp⁴.
The chemical reaction would be written as
2 AsF3<span> + 3 CCl4 = 2 AsCl3 + 3 CCl2F2
</span>
We use the given amounts of the reactants to first find the limiting reactant. Then use the amount of the limiting reactant to proceed to further calculations.
150 g AsF3 ( 1 mol / 131.92 g) = 1.14 mol AsF3
180 g CCl4 (1 mol / 153.82 g) = 1.17 mol CCl4
Therefore, the limiting reactant would be CCl4 since it would be consumed completely. The theoretical yield would be:
1.17 mol CCl4 ( 3 mol CCl2F2 / 3 mol CCl4 ) = 1.17 mol CCl2F2
<span>The element gold has 32 isotopes, ranging from A =173 to A = 204. during alpha decay Au will loose 2 in atomic number and 4 in mass number . it will form a iridium isotope and helium . according to the above statement, the balanced equation for the alpha decay of gold 173 will be given as below.
173 169 4
79 Au-------------->77 Ir + 2He</span>
Answer:
<h2>
The equilibrium constant Kc for this reaction is 19.4760</h2>
Explanation:
The volume of vessel used= 
ml
Initial moles of NO= 
moles
Initial moles of H2= 
moles
Concentration of NO at equilibrium= 
M
Moles of NO at equilibrium= 
=
moles
2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)
<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles
<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles
∴
⇒
![Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BH2O%5D%5E2%5BN2%5D%7D%7B%5BH2%5D%5E2%5BNO%5D%5E2%7D%20%28volume%20of%20vesselin%20litre%29)
<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles
⇒
⇒
