Answer:
Mercury
Explanation:
This is the only metal that exists as a liquid at room temperature
Answer: 1.07×10^-20microlitre
Explanation:
1cm3 = 1000microlitres
1.07×10^-23 cm3 of tungsten = 1.07×10^-23 x 1000 = 1.07×10^-20microlitre
Answer:a) 11.34 g of ethane
can be formed
b)
is the limiting reagent
c) 3.44 g of the excess reagent remains after the reaction is complete
Explanation:
To calculate the moles :
1. ![\text{Moles of} H_2=\frac{4.21}{2}=2.10moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20H_2%3D%5Cfrac%7B4.21%7D%7B2%7D%3D2.10moles)
2. ![\text{Moles of} C_2H_4=\frac{10.6}{28}=0.378moles](https://tex.z-dn.net/?f=%5Ctext%7BMoles%20of%7D%20C_2H_4%3D%5Cfrac%7B10.6%7D%7B28%7D%3D0.378moles)
According to stoichiometry :
1 mole of
require 1 mole of ![H_2](https://tex.z-dn.net/?f=H_2)
Thus 0.378 moles of
will require=
of ![H_2](https://tex.z-dn.net/?f=H_2)
Thus
is the limiting reagent as it limits the formation of product and
is the excess reagent.
moles of
left = (2.10-0.378) = 1.72 moles
mass of
left=![moles\times {\text {Molar mass}}=1.72moles\times 2g/mol=3.44g](https://tex.z-dn.net/?f=moles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D1.72moles%5Ctimes%202g%2Fmol%3D3.44g)
According to stoichiometry :
As 1 mole of
give = 1 mole of ![C_2H_6](https://tex.z-dn.net/?f=C_2H_6)
Thus 0.378 moles of
give =
of ![C_2H_6](https://tex.z-dn.net/?f=C_2H_6)
Mass of ![C_2H_6=moles\times {\text {Molar mass}}=0.378moles\times 30g/mol=11.34g](https://tex.z-dn.net/?f=C_2H_6%3Dmoles%5Ctimes%20%7B%5Ctext%20%7BMolar%20mass%7D%7D%3D0.378moles%5Ctimes%2030g%2Fmol%3D11.34g)
Thus 11.34 g of ethane is formed.
Answer : The solubility of this compound in g/L is
.
Solution : Given,
![K_{sp}=2.42\times 10^{-11}](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D2.42%5Ctimes%2010%5E%7B-11%7D)
Molar mass of
= 114.945g/mole
The balanced equilibrium reaction is,
![MnCO_3\rightleftharpoons Mn^{2+}+CO^{2-}_3](https://tex.z-dn.net/?f=MnCO_3%5Crightleftharpoons%20Mn%5E%7B2%2B%7D%2BCO%5E%7B2-%7D_3)
At equilibrium s s
The expression for solubility constant is,
![K_{sp}=[Mn^{2+}][CO^{2-}_3]](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BMn%5E%7B2%2B%7D%5D%5BCO%5E%7B2-%7D_3%5D)
Now put the given values in this expression, we get
![2.42\times 10^{-11}=(s)(s)\\2.42\times 10^{-11}=s^2\\s=0.4919\times 10^{-5}=4.919\times 10^{-6}moles/L](https://tex.z-dn.net/?f=2.42%5Ctimes%2010%5E%7B-11%7D%3D%28s%29%28s%29%5C%5C2.42%5Ctimes%2010%5E%7B-11%7D%3Ds%5E2%5C%5Cs%3D0.4919%5Ctimes%2010%5E%7B-5%7D%3D4.919%5Ctimes%2010%5E%7B-6%7Dmoles%2FL)
The value of 's' is the molar concentration of manganese ion and carbonate ion.
Now we have to calculate the solubility in terms of g/L multiplying by the Molar mass of the given compound.
![s=4.919\times 10^{-6}moles/L\times 114.945g/mole=565.414\times 10^{-6}g/L](https://tex.z-dn.net/?f=s%3D4.919%5Ctimes%2010%5E%7B-6%7Dmoles%2FL%5Ctimes%20114.945g%2Fmole%3D565.414%5Ctimes%2010%5E%7B-6%7Dg%2FL)
Therefore, the solubility of this compound in g/L is
.