The molarity of the solution is 0.01.
Brainliest?
<u>Answer:</u>
Nitrogen gas be a mineral only, if it is in organic forms.
<u>Explanation:</u>
Most of the forms of organic nitrogen is not be taken by plants, with the exception in the form of small organic molecules. Also plants can promptly take the nitrogen when it is in other forms like ammonia and nitrate.
The microorganisms in the soil converts the organic forms of nitrogen to mineral form when they decompose organic matters and also fresh plant residues. This type of process is called mineralisation.
Answer:
Value of 
for the given redox reaction is 
Explanation:
Redox reaction with states of species:
Reaction quotient for this redox reaction:
![Q_{p}=\frac{[Cr^{3+}]^{2}.P_{Cl_{2}}^{3}}{[H^{+}]^{14}.[Cr_{2}O_{7}^{2-}].[Cl^{-}]^{6}}](https://tex.z-dn.net/?f=Q_%7Bp%7D%3D%5Cfrac%7B%5BCr%5E%7B3%2B%7D%5D%5E%7B2%7D.P_%7BCl_%7B2%7D%7D%5E%7B3%7D%7D%7B%5BH%5E%7B%2B%7D%5D%5E%7B14%7D.%5BCr_%7B2%7DO_%7B7%7D%5E%7B2-%7D%5D.%5BCl%5E%7B-%7D%5D%5E%7B6%7D%7D)
Species inside third braket represent concentration in molarity, P represent pressure in atm and concentration of 
is taken as 1 due to the fact that is a pure liquid.
![pH=-log[H^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH%5E%7B%2B%7D%5D)
So, ![[H^{+}]=10^{-pH}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%3D10%5E%7B-pH%7D)
Plug in all the given values in the equation of :
Answer:
energy required=qnet=87.75kJ
Explanation:
we will do it in three seperate step and then add up those value.
first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.
q 1= m c (T2-T1)
q1 = 36.0 g (4.18 J/gC) (100 - 65 C)
q1 = 5267 J
=5.267kJ
next is to vaporize the sample at 100C
q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol
q2= 81.4 kJ
Finally, heat the steam upto 115C
q3 = m c (T2-T1)
q 3= 36.0 g (2.01 J/gC)(115-100C)
q3 = 1085 J
=1.085kJ
qnet=q1 +q2 +q3
energy required=qnet=87.75kJ
Volume percent<span> or </span>volume/volume percent<span> (v/v%) is used when preparing solutions of liquids. It will have units of volume of the smaller composition substance over the volume of the solution. We calculate as follows:
12.5 mL ethanol = .225 mL ethanol / 1 mL solution ( V )
V = 55.56 mL of the 22.5 % by volume ethanol solution is needed
Hope this answers the question.</span>
