Answer : The name of positive ion present in this sample is, hydronium ion.
Explanation :
For formation of a neutral ionic compound, the charges on cation and anion must be balanced.
The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.
Ion : An ion is formed when an atom looses or gains electron.
When an atom looses electrons, it will form a positive ion known as cation.
When an atom gains electrons, it will form a negative ion known as anion.
As per question, hydronium ion or hydrogen ion is having an oxidation state of +1 called as 
or 
cation and nitrate ion 
is an anion with oxidation state of -1.
Thus they combine and their oxidation states are exchanged and written in simplest whole number ratios to give neutral 
.
Hence, the name of positive ion present in this sample is, hydronium ion.
The answer is B to prevent condensation from forming on the agar surface
Answer:
Equilibrium constant expression for 
:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Capprox%20%5Cfrac%7B%5B%5Cmathrm%7BH_2CO_3%7D%5D%7D%7B%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%5C%2C%20%28aq%29%7D%5Cright%5D%5E%7B2%7D%20%5C%2C%20%5Cleft%5B%5Cmathrm%7BCO_3%7D%5E%7B2-%7D%5Cright%5D%7D)
.
Where
, 
, and 
denote the activities of the three species, and ![[\mathrm{H_2CO_3}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BH_2CO_3%7D%5D)
, ![\left[\mathrm{H^{+}}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%7D%5Cright%5D)
, and ![\left[\mathrm{CO_3}^{2-}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathrm%7BCO_3%7D%5E%7B2-%7D%5Cright%5D)
denote the concentrations of the three species.
Explanation:
<h3>Equilibrium Constant Expression</h3>
The equilibrium constant expression of a (reversible) reaction takes the form a fraction.
Multiply the activity of each product of this reaction to get the numerator.
is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be 
.
Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "
" on the product side of this reaction. 
is equivalent to 
. The species 
appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:
.
That's where the exponent "" in this equilibrium constant expression came from.
Combine these two parts to obtain the equilibrium constant expression:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Cquad%5Cbegin%7Bmatrix%7D%5Cleftarrow%20%5Ctext%7Bfrom%20products%7D%20%5C%5C%5B0.5em%5D%20%5Cleftarrow%20%5Ctext%7Bfrom%20reactants%7D%5Cend%7Bmatrix%7D)
.
<h3 /><h3>Equilibrium Constant of Concentration</h3>
In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "
" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Capprox%20%5Cfrac%7B%5Cleft%5B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%5Cright%5D%7D%7B%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%5C%2C%20%28aq%29%7D%5Cright%5D%5E2%5Ccdot%20%5Cleft%5B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%5Cright%5D%7D)
.
Silver has metallic bonding.
Silver is a very typical and main metal. The negatively charged electrons distribute themselves throughout the entire piece of metal and form non directional bonds between the positive silver ions, which is metallic bonding, and what silver contributes.
Answer:
Mercury has the lowest freezing point for metals, but helium takes the winner for all substances.
Explanation:
